MHT CET 2026 Mathematics Question Paper with Answer and Solution

(B) The given differential equation is $(x^2 + y^2) dy = xy dx$, which can be written as $\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$.
This is a homogeneous differential equation. Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2(1 + v^2)} = \frac{v}{1 + v^2}$.
$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v - v^3}{1 + v^2} = -\frac{v^3}{1 + v^2}$.
Separating variables: $\int \frac{1 + v^2}{v^3} dv = -\int \frac{1}{x} dx$.
$\int (v^{-3} + v^{-1}) dv = -\ln|x| + C$.
$-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C \implies -\frac{x^2}{2y^2} + \ln|y/x| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C \implies \ln|y| - \frac{x^2}{2y^2} = C$.
Given $y(1) = 1$, so $\ln(1) - \frac{1^2}{2(1)^2} = C \implies C = -1/2$.
The equation is $\ln|y| - \frac{x^2}{2y^2} = -1/2$.
Given $y(x_0) = e$, so $\ln(e) - \frac{x_0^2}{2e^2} = -1/2$.
$1 - \frac{x_0^2}{2e^2} = -1/2 \implies \frac{x_0^2}{2e^2} = 3/2 \implies x_0^2 = 3e^2 \implies x_0 = \pm \sqrt{3}e$.

(D) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x - y}{x + y}$.
Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x - vx}{x + vx} = \frac{1 - v}{1 + v}$.
$x \frac{dv}{dx} = \frac{1 - v}{1 + v} - v = \frac{1 - v - v - v^2}{1 + v} = \frac{1 - 2v - v^2}{1 + v}$.
Separating variables: $\int \frac{1 + v}{1 - 2v - v^2} dv = \int \frac{1}{x} dx$.
Let $u = 1 - 2v - v^2$, then $du = (-2 - 2v) dv = -2(1 + v) dv$, so $(1 + v) dv = -\frac{1}{2} du$.
$-\frac{1}{2} \int \frac{1}{u} du = \ln|x| + C \implies -\frac{1}{2} \ln|1 - 2v - v^2| = \ln|x| + C$.
$\ln|1 - 2v - v^2| = -2 \ln|x| + C' \implies 1 - 2(\frac{y}{x}) - (\frac{y}{x})^2 = \frac{K}{x^2}$.
$x^2 - 2xy - y^2 = K$. Since it passes through $(0, 0)$, $K = 0$.
$x^2 - 2xy - y^2 = 0$. This is a homogeneous second-degree equation of the form $Ax^2 + 2Hxy + By^2 = 0$ where $H^2 - AB = (-1)^2 - (1)(-1) = 2 > 0$, representing a pair of straight lines.

(A) Given differential equation: $x \, dy + 2y \, dx = 0$
Rearranging the terms: $x \, dy = -2y \, dx$
Separating variables: $\frac{dy}{2y} = -\frac{dx}{x}$
Integrating both sides: $\int \frac{1}{2y} \, dy = -\int \frac{1}{x} \, dx$
$\frac{1}{2} \ln|y| = -\ln|x| + C$
Multiply by $2$: $\ln|y| = -2\ln|x| + 2C$
$\ln|y| = \ln|x^{-2}| + C_1$ (where $C_1 = 2C$)
$y = e^{C_1} \cdot x^{-2} \implies y = \frac{K}{x^2}$ (where $K = e^{C_1}$)
Given $x = 2$ and $y = 1$: $1 = \frac{K}{2^2} \implies K = 4$
Thus, the particular solution is $y = \frac{4}{x^2}$, which simplifies to $x^2 y = 4$.

(C) Let $S = 4\pi r^2$ be the surface area. Given $\frac{dS}{dt} = kS$.
Integrating, $\ln S = kt + C$, so $S(t) = S_0 e^{kt}$.
At $t = 0$, $r = 2 \text{ cm}$, so $S_0 = 4\pi(2)^2 = 16\pi \text{ cm}^2$.
At $t = 5$, $r = 7 \text{ cm}$, so $S(5) = 4\pi(7)^2 = 196\pi \text{ cm}^2$.
$196\pi = 16\pi e^{5k} \implies e^{5k} = \frac{196}{16} = 12.25$.
$e^k = (12.25)^{1/5}$.
At $t = 12$, $S(12) = 16\pi (e^k)^{12} = 16\pi (12.25)^{12/5}$.
$(12.25)^{2.4} \approx 12.25^{2} \times 12.25^{0.4} \approx 150.0625 \times 3.24 \approx 49$.
More precisely, $S(12) = 16\pi (12.25)^{2.4} \approx 16 \times 3.14159 \times 49 \approx 2463.01 \text{ cm}^2$.
Rounding to the nearest option, $S(12) = 2464 \text{ cm}^2$.

(A) Step $1$: Differentiate $(y - a)^2 = 4(x - b)$ with respect to $x$: $2(y - a) \frac{dy}{dx} = 4$, which simplifies to $(y - a) \frac{dy}{dx} = 2$.
Step $2$: Differentiate again with respect to $x$: $\frac{dy}{dx} \cdot \frac{dy}{dx} + (y - a) \frac{d^2y}{dx^2} = 0$.
Step $3$: From Step $1$, $(y - a) = \frac{2}{dy/dx}$. Substitute this into the equation from Step $2$: $(\frac{dy}{dx})^2 + \frac{2}{dy/dx} \cdot \frac{d^2y}{dx^2} = 0$.
Step $4$: Multiply by $\frac{dy}{dx}$ to clear the fraction: $(\frac{dy}{dx})^3 + 2 \frac{d^2y}{dx^2} = 0$.
Step $5$: The highest order derivative is $\frac{d^2y}{dx^2}$, and its power is $1$. Thus, the degree is $1$.

(A) Step $1$: The equation of the line in normal form is $x \cos \alpha + y \sin \alpha = p$.
Step $2$: Differentiate with respect to $x$: $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$.
Step $3$: This implies $\frac{dy}{dx} = -\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha$.
Step $4$: Differentiate again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(-\cot \alpha) = 0$ (since $\alpha$ is a constant).
Step $5$: Thus, the differential equation is $\frac{d^2y}{dx^2} = 0$.

(B) The general equation of a parabola with its axis along the $y$-axis is $y = ax^2 + b$.
Since there are two arbitrary constants $a$ and $b$, we differentiate twice.
Differentiating with respect to $x$: $\frac{dy}{dx} = 2ax$.
Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = 2a$.
From the first derivative, $a = \frac{1}{2x} \frac{dy}{dx}$.
Substitute $a$ into the second derivative: $\frac{d^2y}{dx^2} = 2 \left( \frac{1}{2x} \frac{dy}{dx} \right) = \frac{1}{x} \frac{dy}{dx}$.
Rearranging gives $x \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.

(A) Step $1$: Rewrite the equation to remove the fractional exponent: $3 - (\frac{d^3y}{dx^3})^{2/3} = (\frac{dy}{dx})^5 \implies 3 - (\frac{dy}{dx})^5 = (\frac{d^3y}{dx^3})^{2/3}$.
Step $2$: Cube both sides to eliminate the fractional power: $(3 - (\frac{dy}{dx})^5)^3 = (\frac{d^3y}{dx^3})^2$.
Step $3$: The highest order derivative present is $\frac{d^3y}{dx^3}$, so the order is $3$.
Step $4$: The power of the highest order derivative after making the equation free from radicals and fractions is $2$, so the degree is $2$.

(B) Step $1$: Given the differential equation $\sqrt{2 + (\frac{d^2y}{dx^2})^3} = (\frac{d^3y}{dx^3})^{5/2}$.
Step $2$: Square both sides to remove the square root: $2 + (\frac{d^2y}{dx^2})^3 = (\frac{d^3y}{dx^3})^5$.
Step $3$: The order of a differential equation is the highest derivative present. Here, the highest derivative is $\frac{d^3y}{dx^3}$, so the order is $3$.
Step $4$: The degree is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. The power of $\frac{d^3y}{dx^3}$ is $5$. Thus, the degree is $5$.

(C) Step $1$: The order of a differential equation is the order of the highest derivative present in the equation. Here, the highest derivative is $\frac{d^3y}{dx^3}$, so the order is $3$.
Step $2$: The degree of a differential equation is the power of the highest derivative, provided the equation is a polynomial in terms of its derivatives.
Step $3$: In this equation, the term $\cos(\frac{d^2y}{dx^2})$ involves a transcendental function of a derivative, meaning it cannot be expressed as a polynomial in terms of its derivatives.
Step $4$: Therefore, the degree of this differential equation is not defined.

(A) Step $1$: Write the given equation: $\sqrt{1 + \frac{1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$.
Step $2$: Simplify the expression inside the square root: $\sqrt{\frac{(dy/dx)^2 + 1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$.
Step $3$: This becomes $\frac{\sqrt{(dy/dx)^2 + 1}}{|dy/dx|} = (d^2y/dx^2)^{3/2}$.
Step $4$: Square both sides to remove the square root on the right: $\frac{(dy/dx)^2 + 1}{(dy/dx)^2} = (d^2y/dx^2)^3$.
Step $5$: Multiply by $(dy/dx)^2$ to clear the fraction: $(dy/dx)^2 + 1 = (d^2y/dx^2)^3 \cdot (dy/dx)^2$.
Step $6$: The highest order derivative is $d^2y/dx^2$, so the order is $2$. The power of the highest order derivative is $3$, so the degree is $3$.

(B) Step $1$: The given differential equation is $(\frac{d^3y}{dx^3})^{2/3} - 3 \frac{d^2y}{dx^2} + 5 \frac{dy}{dx} + 4 = 0$.
Step $2$: To find the degree, the equation must be a polynomial in terms of its derivatives. We eliminate the fractional exponent by raising both sides to the power of $3$: $(\frac{d^3y}{dx^3})^2 = (3 \frac{d^2y}{dx^2} - 5 \frac{dy}{dx} - 4)^3$.
Step $3$: The highest order derivative present is $\frac{d^3y}{dx^3}$, so the order is $3$.
Step $4$: The power to which the highest order derivative is raised after making the equation a polynomial is $2$, so the degree is $2$.

(B) Step $1$: Given equation is $\sqrt{1 + \frac{1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$.
Step $2$: Simplify the expression inside the square root: $\sqrt{\frac{(dy/dx)^2 + 1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$.
Step $3$: This can be written as $\frac{\sqrt{1 + (dy/dx)^2}}{|dy/dx|} = (d^2y/dx^2)^{3/2}$.
Step $4$: Square both sides to remove the square root on the right side: $\frac{1 + (dy/dx)^2}{(dy/dx)^2} = (d^2y/dx^2)^3$.
Step $5$: Multiply by $(dy/dx)^2$ to clear the fraction: $1 + (dy/dx)^2 = (dy/dx)^2 \cdot (d^2y/dx^2)^3$.
Step $6$: The highest order derivative present is $d^2y/dx^2$, so the order is $2$.
Step $7$: The power of the highest order derivative after making the equation a polynomial in derivatives is $3$, so the degree is $3$.

(A) Step $1$: Simplify the given equation. $\sqrt{\frac{(\frac{dy}{dx})^2 + 1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$.
Step $2$: Square both sides to remove the square root: $\frac{(\frac{dy}{dx})^2 + 1}{(\frac{dy}{dx})^2} = (\frac{d^2y}{dx^2})^3$.
Step $3$: Multiply by $(\frac{dy}{dx})^2$ to clear the fraction: $(\frac{dy}{dx})^2 + 1 = (\frac{d^2y}{dx^2})^3 (\frac{dy}{dx})^2$.
Step $4$: The highest order derivative present is $\frac{d^2y}{dx^2}$, so the order is $2$.
Step $5$: The power of the highest order derivative after making the equation a polynomial in derivatives is $3$, so the degree is $3$.

(B) Step $1$: Write the given differential equation: $(\frac{d^3y}{dx^3})^{\frac{2}{3}} = 3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4$.
Step $2$: To find the degree, the equation must be a polynomial in terms of derivatives. Remove the fractional exponent by raising both sides to the power of $3$: $((\frac{d^3y}{dx^3})^{\frac{2}{3}})^3 = (3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4)^3$.
Step $3$: This simplifies to $(\frac{d^3y}{dx^3})^2 = (3\frac{d^2y}{dx^2} - 5\frac{dy}{dx} - 4)^3$.
Step $4$: The order is the highest derivative present, which is $3$. The degree is the power of the highest derivative after rationalizing, which is $2$. Thus, the order is $3$ and the degree is $2$.

(B) Given equation: $\sqrt{1 + \frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$
Simplify the left side: $\sqrt{\frac{(\frac{dy}{dx})^2 + 1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$
$\frac{\sqrt{(\frac{dy}{dx})^2 + 1}}{|\frac{dy}{dx}|} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}$
Square both sides to remove the fractional exponent: $\frac{(\frac{dy}{dx})^2 + 1}{(\frac{dy}{dx})^2} = (\frac{d^2y}{dx^2})^3$
Multiply by $(\frac{dy}{dx})^2$: $(\frac{dy}{dx})^2 + 1 = (\frac{d^2y}{dx^2})^3 (\frac{dy}{dx})^2$
The highest order derivative present is $\frac{d^2y}{dx^2}$, so the order is $2$.
The power of the highest order derivative after making the equation a polynomial in derivatives is $3$, so the degree is $3$.
Thus, the order and degree are $2$ and $3$ respectively.

(B) Given $\vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0}$, we have $\vec{b} = -\vec{a} \times (\vec{a} \times \vec{c})$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$, we get $\vec{b} = -[(\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}] = (\vec{a} \cdot \vec{a})\vec{c} - (\vec{a} \cdot \vec{c})\vec{a}$.
Since $|\vec{a}| = 1$, $|\vec{a} \cdot \vec{a}| = 1^2 = 1$. Thus, $\vec{b} = \vec{c} - (\vec{a} \cdot \vec{c})\vec{a}$.
Taking the magnitude squared on both sides: $|\vec{b}|^2 = |\vec{c} - (\vec{a} \cdot \vec{c})\vec{a}|^2 = |\vec{c}|^2 + (\vec{a} \cdot \vec{c})^2 |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c})(\vec{c} \cdot \vec{a}) = |\vec{c}|^2 - (\vec{a} \cdot \vec{c})^2$.
Given $|\vec{b}| = 1$ and $|\vec{c}| = 2$, we have $1^2 = 2^2 - (\vec{a} \cdot \vec{c})^2$, so $(\vec{a} \cdot \vec{c})^2 = 3$.
Since $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos \theta = 1 \cdot 2 \cos \theta = 2 \cos \theta$, we have $(2 \cos \theta)^2 = 3$, which means $4 \cos^2 \theta = 3$.
$\cos^2 \theta = \frac{3}{4} \implies \cos \theta = \frac{\sqrt{3}}{2}$ (for acute angle $\theta$).
Therefore, $\theta = \frac{\pi}{6}$.

(C) Using the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
Given $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b}}{2}$, we have $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b}$.
Rearranging the terms: $(\vec{a} \cdot \vec{c} - \frac{1}{2})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = 0$.
Since $\vec{b}$ and $\vec{c}$ are non-coplanar, they are linearly independent. Therefore, their coefficients must be zero.
Thus, $\vec{a} \cdot \vec{b} = 0$.
Since $\vec{a}$ and $\vec{b}$ are non-zero vectors and their dot product is $0$, the angle between them is $\frac{\pi}{2}$.

(D) Given $\vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = \vec{0}$.
Using the vector triple product formula $\vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}$.
Since $|\vec{a}| = 1$, we have $\vec{a} \cdot \vec{a} = 1$.
So, $(\vec{a} \cdot \vec{c})\vec{a} - \vec{c} + \vec{b} = \vec{0}$, which implies $\vec{b} = \vec{c} - (\vec{a} \cdot \vec{c})\vec{a}$.
Taking the magnitude squared on both sides: $|\vec{b}|^2 = |\vec{c}|^2 + (\vec{a} \cdot \vec{c})^2 |\vec{a}|^2 - 2(\vec{a} \cdot \vec{c})(\vec{c} \cdot \vec{a})$.
Since $|\vec{b}| = 1$, $|\vec{c}| = 2$, and $|\vec{a}| = 1$, we get $1 = 4 + (\vec{a} \cdot \vec{c})^2 - 2(\vec{a} \cdot \vec{c})^2$.
$1 = 4 - (\vec{a} \cdot \vec{c})^2$, so $(\vec{a} \cdot \vec{c})^2 = 3$.
Thus, $|\vec{a}||\vec{c}| \cos \theta = \pm \sqrt{3}$, where $\theta$ is the angle between $\vec{a}$ and $\vec{c}$.
$1 \cdot 2 \cdot \cos \theta = \sqrt{3} \implies \cos \theta = \frac{\sqrt{3}}{2}$.
Therefore, $\theta = \frac{\pi}{6}$.

(D) Let $V = [\vec{a} \vec{b} \vec{c}]$. By definition, $\vec{a} \cdot \vec{p} = \vec{a} \cdot \frac{\vec{b} \times \vec{c}}{V} = \frac{[\vec{a} \vec{b} \vec{c}]}{V} = 1$.
Similarly, $\vec{b} \cdot \vec{q} = 1$ and $\vec{c} \cdot \vec{r} = 1$.
Also, $\vec{b} \cdot \vec{p} = 0$, $\vec{c} \cdot \vec{p} = 0$, $\vec{c} \cdot \vec{q} = 0$, $\vec{a} \cdot \vec{q} = 0$, $\vec{a} \cdot \vec{r} = 0$, and $\vec{b} \cdot \vec{r} = 0$ because the dot product of a vector with a cross product containing itself is zero.
Now, expand the expression: $(\vec{a} + \vec{b}) \cdot \vec{p} + (\vec{b} + \vec{c}) \cdot \vec{q} + (\vec{c} + \vec{a}) \cdot \vec{r} = (\vec{a} \cdot \vec{p} + \vec{b} \cdot \vec{p}) + (\vec{b} \cdot \vec{q} + \vec{c} \cdot \vec{q}) + (\vec{c} \cdot \vec{r} + \vec{a} \cdot \vec{r})$.
Substituting the values: $(1 + 0) + (1 + 0) + (1 + 0) = 3$.

(B) Three vectors are coplanar if their scalar triple product is zero.
Let the vectors be $\vec{u} = a\hat{i} + a\hat{j} + c\hat{k}$, $\vec{v} = \hat{i} + 0\hat{j} + \hat{k}$, and $\vec{w} = c\hat{i} + c\hat{j} + b\hat{k}$.
The condition for coplanarity is $\det(\vec{u}, \vec{v}, \vec{w}) = 0$.
$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$.
Expanding along the second row: $-1(ab - c^2) + 0 - 1(ac - ac) = 0$.
$-(ab - c^2) = 0$.
$c^2 = ab$.
$c = \sqrt{ab}$.
Thus, $c$ is the geometric mean of $a$ and $b$.

(A) Step $1$: Calculate the volume of the parallelepiped using the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$.
$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(4 - 6) + \hat{k}(1 - (-9)) = 14\hat{i} + 2\hat{j} + 10\hat{k}$.
Volume $V = |\vec{a} \cdot (\vec{b} \times \vec{c})| = |(-2)(14) + (5)(2) + (3)(10)| = |-28 + 10 + 30| = 12$.
Step $2$: Calculate the area of the base formed by $\vec{b}$ and $\vec{c}$, which is $|\vec{b} \times \vec{c}|$.
$|\vec{b} \times \vec{c}| = \sqrt{14^2 + 2^2 + 10^2} = \sqrt{196 + 4 + 100} = \sqrt{300} = 10\sqrt{3}$.
Step $3$: The altitude $h$ is given by $V / \text{Area} = 12 / (10\sqrt{3}) = 6 / (5\sqrt{3}) = (6\sqrt{3}) / 15 = 2\sqrt{3} / 5$.

(D) The function $f(x)$ is the scalar triple product $[\vec{a} \vec{b} \vec{c}]$, given by the determinant:
$f(x) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$
Expanding along the first row:
$f(x) = x(x^2 - 2) + 2(-2x + 7) + 3(4 - 7x)$
$f(x) = x^3 - 2x - 4x + 14 + 12 - 21x = x^3 - 27x + 26$
To find local maxima, find $f'(x) = 3x^2 - 27$. Setting $f'(x) = 0$ gives $x^2 = 9$, so $x = \pm 3$.
$f''(x) = 6x$. For local maxima, $f''(x) < 0$, so $x_0 = -3$.
Now calculate $\vec{a} \cdot \vec{c}$ at $x = -3$:
$\vec{a} = -3\hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{c} = 7\hat{i} - 2\hat{j} - 3\hat{k}$.
$\vec{a} \cdot \vec{c} = (-3)(7) + (-2)(-2) + (3)(-3) = -21 + 4 - 9 = -26$.

(B) The scalar triple product is defined as $[\vec{a} \quad \vec{a} + \vec{b} \quad \vec{a} \times \vec{b}] = \vec{a} \cdot ((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b}))$.
Using the vector triple product identity $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$, we expand $(\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})$:
$(\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b}) = \vec{a} \times (\vec{a} \times \vec{b}) + \vec{b} \times (\vec{a} \times \vec{b})$
$= ((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) + ((\vec{b} \cdot \vec{b})\vec{a} - (\vec{b} \cdot \vec{a})\vec{b})$
$= (\vec{a} \cdot \vec{b})\vec{a} - |\vec{a}|^2 \vec{b} + |\vec{b}|^2 \vec{a} - (\vec{a} \cdot \vec{b})\vec{b}$
Now, take the dot product with $\vec{a}$:
$[\vec{a} \quad \vec{a} + \vec{b} \quad \vec{a} \times \vec{b}] = \vec{a} \cdot [(\vec{a} \cdot \vec{b})\vec{a} - |\vec{a}|^2 \vec{b} + |\vec{b}|^2 \vec{a} - (\vec{a} \cdot \vec{b})\vec{b}]$
$= (\vec{a} \cdot \vec{b})(\vec{a} \cdot \vec{a}) - |\vec{a}|^2 (\vec{a} \cdot \vec{b}) + |\vec{b}|^2 (\vec{a} \cdot \vec{a}) - (\vec{a} \cdot \vec{b})(\vec{a} \cdot \vec{b})$
$= (8)(16) - (16)(8) + (9)(16) - (8)(8)$
$= 0 + 144 - 64 = 80$.

(D) Since $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, the scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}] = 0$.
This is equivalent to the determinant of the components being zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$.
Expanding along the first row:
$1((-1)(-1) - (2)(x-2)) - 1((1)(-1) - (2)(x)) + 1((1)(x-2) - (-1)(x)) = 0$.
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$.
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$.
$4 + 2x = 0$.
$2x = -4$.
$x = -2$.

(A) The volume of a tetrahedron with vertices $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ is given by $V = \frac{1}{6} |(\vec{b}-\vec{a}) \cdot ((\vec{c}-\vec{a}) \times (\vec{d}-\vec{a}))|$.
Let $\vec{a} = (5, -1, 1)$, $\vec{b} = (7, -4, p)$, $\vec{c} = (1, -6, 10)$, $\vec{d} = (-1, -3, 7)$.
$\vec{b}-\vec{a} = (2, -3, p-1)$, $\vec{c}-\vec{a} = (-4, -5, 9)$, $\vec{d}-\vec{a} = (-6, -2, 6)$.
Volume $V = \frac{1}{6} |\det \begin{pmatrix} 2 & -3 & p-1 \\ -4 & -5 & 9 \\ -6 & -2 & 6 \end{pmatrix}| = 11$.
$|2(-30 + 18) + 3(-24 + 54) + (p-1)(8 - 30)| = 66$.
$|2(-12) + 3(30) + (p-1)(-22)| = 66$.
$|-24 + 90 - 22p + 22| = 66$.
$|88 - 22p| = 66$.
Case $1$: $88 - 22p = 66 \implies 22p = 22 \implies p = 1$.
Case $2$: $88 - 22p = -66 \implies 22p = 154 \implies p = 7$.
Thus, one of the values of $p$ is $1$.

(A) Given, the volume of the parallelepiped is $|[\vec{a} \vec{b} \vec{c}]| = 3$.
The volume of a tetrahedron with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by $V = \frac{1}{6} |[\vec{u} \vec{v} \vec{w}]|$.
Here, $\vec{u} = \vec{a} \times \vec{b}$, $\vec{v} = 2(\vec{a} \times \vec{c})$, $\vec{w} = 2(\vec{b} \times \vec{c})$.
Thus, $[\vec{u} \vec{v} \vec{w}] = [(\vec{a} \times \vec{b}) \ (2(\vec{a} \times \vec{c})) \ (2(\vec{b} \times \vec{c}))] = 4 [(\vec{a} \times \vec{b}) \ (\vec{a} \times \vec{c}) \ (\vec{b} \times \vec{c})]$.
Using the identity $[(\vec{a} \times \vec{b}) \ (\vec{b} \times \vec{c}) \ (\vec{c} \times \vec{a})] = [\vec{a} \vec{b} \vec{c}]^2$, we have $[(\vec{a} \times \vec{b}) \ (\vec{a} \times \vec{c}) \ (\vec{b} \times \vec{c})] = [\vec{a} \vec{b} \vec{c}]^2$.
So, $[\vec{u} \vec{v} \vec{w}] = 4 [\vec{a} \vec{b} \vec{c}]^2 = 4 \times (3)^2 = 4 \times 9 = 36$.
The volume of the tetrahedron is $V = \frac{1}{6} \times 36 = 6 \text{ cubic units}$.

(A) The determinant $\left| \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right|$ represents the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$.
Since $\vec{a}$ is a unit vector perpendicular to both $\vec{b}$ and $\vec{c}$, $\vec{a} = \pm \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}$.
Thus, $\vec{a} \cdot (\vec{b} \times \vec{c}) = \pm \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} \cdot (\vec{b} \times \vec{c}) = \pm |\vec{b} \times \vec{c}|$.
Squaring both sides, we get the value as $|\vec{b} \times \vec{c}|^2$.
Using the formula $|\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}| \sin(\theta)$, where $\theta = \frac{\pi}{3}$.
$|\vec{b} \times \vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 \sin^2(\frac{\pi}{3}) = |\vec{b}|^2|\vec{c}|^2 (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}|\vec{b}|^2|\vec{c}|^2$.

(B) Given $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$, we can write $(\vec{r} - \vec{c}) \times \vec{b} = 0$. This implies $\vec{r} - \vec{c} = t\vec{b}$ for some scalar $t$, so $\vec{r} = \vec{c} + t\vec{b}$.
Substituting $\vec{r}$ into $\vec{r} \cdot \vec{a} = 0$, we get $(\vec{c} + t\vec{b}) \cdot \vec{a} = 0$, which means $\vec{c} \cdot \vec{a} + t(\vec{b} \cdot \vec{a}) = 0$.
Calculate $\vec{c} \cdot \vec{a} = (4)(2) + (-3)(0) + (7)(1) = 8 + 0 + 7 = 15$.
Calculate $\vec{b} \cdot \vec{a} = (1)(2) + (1)(0) + (1)(1) = 2 + 0 + 1 = 3$.
Thus, $15 + 3t = 0$, which gives $t = -5$.
Now, $\vec{r} = \vec{c} - 5\vec{b} = (4\hat{i} - 3\hat{j} + 7\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}) = -\hat{i} - 8\hat{j} + 2\hat{k}$.
Finally, $\vec{r} \cdot \vec{c} = (-1)(4) + (-8)(-3) + (2)(7) = -4 + 24 + 14 = 34$.

(B) The line passing through $P(1, 0, -3)$ and parallel to the given line has the equation $\frac{x - 1}{2} = \frac{y - 0}{3} = \frac{z + 3}{-6} = r$.
Any point on this line is $(2r + 1, 3r, -6r - 3)$.
For this point to lie on the plane $x - y - z = 9$, we substitute the coordinates:
$(2r + 1) - (3r) - (-6r - 3) = 9$.
$2r + 1 - 3r + 6r + 3 = 9$.
$5r + 4 = 9 \implies 5r = 5 \implies r = 1$.
The point of intersection is $(2(1) + 1, 3(1), -6(1) - 3) = (3, 3, -9)$.
The distance $d$ between $(1, 0, -3)$ and $(3, 3, -9)$ is $\sqrt{(3 - 1)^2 + (3 - 0)^2 + (-9 - (-3))^2}$.
$d = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \text{ units}$.

(A) The line is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{b} = \hat{i} + \hat{j} + \hat{k}$.
The plane is given by $\vec{r} \cdot \vec{n} = d$, where $\vec{n} = 2\hat{i} - \hat{j} + \hat{k}$.
The angle $\theta$ between a line and a plane is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Calculate the dot product: $\vec{b} \cdot \vec{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$.
Calculate the magnitudes: $|\vec{b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
Substitute into the formula: $\sin \theta = \frac{|2|}{\sqrt{3} \cdot \sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$.
Therefore, $\theta = \sin^{-1} \left( \frac{\sqrt{2}}{3} \right)$.

(B) The formula for the sine of the angle $\theta$ between a line with direction vector $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and a plane with normal vector $\vec{n} = 2\hat{i} + p\hat{j} + \hat{k}$ is $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\sin \theta = \frac{\sqrt{2}}{3}$, $\vec{b} \cdot \vec{n} = (1)(2) + (1)(p) + (1)(1) = p + 3$.
Magnitudes are $|\vec{b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ and $|\vec{n}| = \sqrt{2^2 + p^2 + 1^2} = \sqrt{p^2 + 5}$.
Substituting these into the formula: $\frac{\sqrt{2}}{3} = \frac{|p + 3|}{\sqrt{3} \sqrt{p^2 + 5}}$.
Squaring both sides: $\frac{2}{9} = \frac{(p + 3)^2}{3(p^2 + 5)}$.
Simplifying: $\frac{2}{3} = \frac{p^2 + 6p + 9}{p^2 + 5}$.
$2(p^2 + 5) = 3(p^2 + 6p + 9) \implies 2p^2 + 10 = 3p^2 + 18p + 27$.
$p^2 + 18p + 17 = 0$.
$(p + 1)(p + 17) = 0$.
Thus, $p = -1$ or $p = -17$.

(B) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$.
Given plane: $3x + 4y - 12z + 13 = 0$.
Distance of $(1, 1, \mu)$ from the plane: $d_1 = \frac{|3(1) + 4(1) - 12(\mu) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12\mu|}{13}$.
Distance of $(-3, 0, 1)$ from the plane: $d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.
Since the points are equidistant, $d_1 = d_2$, so $\frac{|20 - 12\mu|}{13} = \frac{8}{13}$.
$|20 - 12\mu| = 8$.
Case $1$: $20 - 12\mu = 8 \implies 12\mu = 12 \implies \mu = 1$.
Case $2$: $20 - 12\mu = -8 \implies 12\mu = 28 \implies \mu = \frac{28}{12} = \frac{7}{3}$.
Thus, the values of $\mu$ are $1, \frac{7}{3}$.

(A) The normal vector $\vec{n}$ to the plane is given by the cross product of the parallel vectors $\vec{b}$ and $\vec{c}$.
$\vec{n} = \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 3) - \hat{j}(4 - 3) + \hat{k}(4 + 1) = -4\hat{i} - \hat{j} + 5\hat{k}$.
The vector equation of a plane passing through point $\vec{a}$ with normal $\vec{n}$ is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Here, $\vec{a} = -2\hat{i} + 7\hat{j} + 5\hat{k}$.
$\vec{a} \cdot \vec{n} = (-2\hat{i} + 7\hat{j} + 5\hat{k}) \cdot (-4\hat{i} - \hat{j} + 5\hat{k}) = (-2)(-4) + (7)(-1) + (5)(5) = 8 - 7 + 25 = 26$.
Thus, the equation is $\vec{r} \cdot (-4\hat{i} - \hat{j} + 5\hat{k}) = 26$.

(A) Step $1$: Identify the coordinates of the points $A, B, C$ where the plane intersects the axes.
For $x$-axis, $y=0, z=0 \implies \frac{x}{2} = 1 \implies x=2$. So, $A = (2, 0, 0)$.
For $y$-axis, $x=0, z=0 \implies \frac{y}{3} = 1 \implies y=3$. So, $B = (0, 3, 0)$.
For $z$-axis, $x=0, y=0 \implies \frac{z}{4} = 1 \implies z=4$. So, $C = (0, 0, 4)$.
Step $2$: Find vectors $\vec{AB}$ and $\vec{AC}$.
$\vec{AB} = B - A = (0-2, 3-0, 0-0) = (-2, 3, 0)$.
$\vec{AC} = C - A = (0-2, 0-0, 4-0) = (-2, 0, 4)$.
Step $3$: Calculate the cross product $\vec{AB} \times \vec{AC}$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ -2 & 0 & 4 \end{vmatrix} = \hat{i}(12-0) - \hat{j}(-8-0) + \hat{k}(0 - (-6)) = 12\hat{i} + 8\hat{j} + 6\hat{k}$.
Step $4$: Calculate the area of triangle $ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Area $= \frac{1}{2} \sqrt{12^2 + 8^2 + 6^2} = \frac{1}{2} \sqrt{144 + 64 + 36} = \frac{1}{2} \sqrt{244} = \frac{1}{2} \sqrt{4 \times 61} = \frac{2\sqrt{61}}{2} = \sqrt{61}$ sq. units.

(B) The line $L$ passes through point $B(2, 3, -2)$ and has direction vector $\vec{v} = -3\hat{i} + 2\hat{j} + \hat{k}$.
The vector $\vec{AB} = (2-3)\hat{i} + (3-7)\hat{j} + (-2 - (-7))\hat{k} = -\hat{i} - 4\hat{j} + 5\hat{k}$.
The normal vector $\vec{n}$ to the plane is $\vec{n} = \vec{AB} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -4 & 5 \\ -3 & 2 & 1 \end{vmatrix} = \hat{i}(-4-10) - \hat{j}(-1+15) + \hat{k}(-2-12) = -14\hat{i} - 14\hat{j} - 14\hat{k}$.
We can take $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
The equation of the plane is $1(x-3) + 1(y-7) + 1(z+7) = 0$, which simplifies to $x + y + z - 3 = 0$.
The distance $d$ from the origin $(0, 0, 0)$ is $d = \frac{|0+0+0-3|}{\sqrt{1^2+1^2+1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Therefore, $d^2 = 3$.

(A) The equation of the plane is $2x - 3y - 6z = 49$. The normal vector to the plane is $\vec{n} = 2\hat{i} - 3\hat{j} - 6\hat{k}$.
The line passing through the origin $(0, 0, 0)$ and perpendicular to the plane has the direction ratios of the normal vector. Thus, the equation of the line is $\frac{x}{2} = \frac{y}{-3} = \frac{z}{-6} = k$.
Any point on this line is $(2k, -3k, -6k)$.
Since this point lies on the plane, substitute it into the plane equation: $2(2k) - 3(-3k) - 6(-6k) = 49$.
$4k + 9k + 36k = 49 \implies 49k = 49 \implies k = 1$.
Substituting $k = 1$ into the point coordinates, we get $(2(1), -3(1), -6(1)) = (2, -3, -6)$.

(A) The equation of a plane passing through the intersection of planes $P_1: 2x - y + z - 3 = 0$ and $P_2: 4x - 3y + 5z + 9 = 0$ is given by $(2x - y + z - 3) + \lambda(4x - 3y + 5z + 9) = 0$.
Rearranging the terms: $(2 + 4\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda)z + (-3 + 9\lambda) = 0$.
The normal vector to this plane is $\vec{n} = (2 + 4\lambda)\hat{i} + (-1 - 3\lambda)\hat{j} + (1 + 5\lambda)\hat{k}$.
Since the plane is parallel to the line with direction vector $\vec{v} = 2\hat{i} + 4\hat{j} + 5\hat{k}$, the normal vector $\vec{n}$ must be perpendicular to $\vec{v}$, so $\vec{n} \cdot \vec{v} = 0$.
$2(2 + 4\lambda) + 4(-1 - 3\lambda) + 5(1 + 5\lambda) = 0$.
$4 + 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$.
$21\lambda + 5 = 0 \implies \lambda = -5/21$.
Substituting $\lambda = -5/21$ into the plane equation: $(2 - 20/21)x + (-1 + 15/21)y + (1 - 25/21)z + (-3 - 45/21) = 0$.
$(42 - 20)/21 x + (-21 + 15)/21 y + (21 - 25)/21 z + (-63 - 45)/21 = 0$.
$22x - 6y - 4z - 108 = 0$.
Dividing by $2$: $11x - 3y - 2z - 54 = 0$.

(B) The vector equation of a plane at a distance $d$ from the origin and normal to a vector $\vec{n}$ is given by $\vec{r} \cdot \hat{n} = d$, where $\hat{n} = \frac{\vec{n}}{|\vec{n}|}$ is the unit normal vector.
Given $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $d = 5$.
Calculate the magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
The unit normal vector is $\hat{n} = \frac{2\hat{i} + \hat{j} - 2\hat{k}}{3}$.
The equation of the plane is $\vec{r} \cdot \left( \frac{2\hat{i} + \hat{j} - 2\hat{k}}{3} \right) = 5$.
Multiplying both sides by $3$, we get $\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 15$.

(B) The lines pass through points $A(1, -1, 0)$ and $B(-1, -1, 0)$. The direction vectors are $\vec{v_1} = 2\hat{i} + \lambda\hat{j} + 2\hat{k}$ and $\vec{v_2} = 5\hat{i} + 2\hat{j} + \lambda\hat{k}$.
Since the lines lie in the same plane, the vector $\vec{AB} = -2\hat{i} + 0\hat{j} + 0\hat{k}$ must be coplanar with $\vec{v_1}$ and $\vec{v_2}$. Thus, the scalar triple product $[\vec{AB}, \vec{v_1}, \vec{v_2}] = 0$.
$\begin{vmatrix} -2 & 0 & 0 \\ 2 & \lambda & 2 \\ 5 & 2 & \lambda \end{vmatrix} = -2(\lambda^2 - 4) = 0 \implies \lambda^2 = 4 \implies \lambda = \pm 2$.
The normal to the plane is $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & \lambda & 2 \\ 5 & 2 & \lambda \end{vmatrix} = (\lambda^2 - 4)\hat{i} - (2\lambda - 10)\hat{j} + (4 - 5\lambda)\hat{k}$.
For $\lambda = 2$, $\vec{n} = 0\hat{i} + 6\hat{j} - 6\hat{k} = 6(\hat{j} - \hat{k})$. The plane is $0(x-1) + 1(y+1) - 1(z-0) = 0 \implies y - z + 1 = 0$.
For $\lambda = -2$, $\vec{n} = 0\hat{i} + 14\hat{j} + 14\hat{k} = 14(\hat{j} + \hat{k})$. The plane is $0(x-1) + 1(y+1) + 1(z-0) = 0 \implies y + z + 1 = 0$.
Combining these, the equation is $y \pm z + 1 = 0$.

(A) Step $1$: The normal vectors of the given planes are $\vec{n_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 3\hat{i} + 3\hat{j} + 2\hat{k}$.
Step $2$: The normal vector $\vec{n}$ of the required plane is perpendicular to both $\vec{n_1}$ and $\vec{n_2}$, so $\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{vmatrix} = \hat{i}(4 - 6) - \hat{j}(2 - 6) + \hat{k}(3 - 6) = -2\hat{i} + 4\hat{j} - 3\hat{k}$.
Step $3$: The equation of the plane passing through $(x_0, y_0, z_0) = (1, 2, 1)$ with normal $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Step $4$: Substituting the values: $-2(x - 1) + 4(y - 2) - 3(z - 1) = 0 \implies -2x + 2 + 4y - 8 - 3z + 3 = 0 \implies -2x + 4y - 3z - 3 = 0$, which is $2x - 4y + 3z + 3 = 0$.

(D) Let the equation of the plane be $a(x-1) + by + cz = 0$, which simplifies to $ax + by + cz = a$. Since it passes through $(0, 1, 0)$, we have $a(0) + b(1) + c(0) = a$, so $b = a$. The normal vector is $\vec{n_1} = (a, a, c)$.
The normal to the plane $2x + 3y = 7$ is $\vec{n_2} = (2, 3, 0)$.
The angle $\theta = 45^{\circ}$ between the planes is given by $\cos(45^{\circ}) = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\frac{1}{\sqrt{2}} = \frac{|2a + 3a + 0|}{\sqrt{a^2 + a^2 + c^2} \sqrt{2^2 + 3^2}} = \frac{|5a|}{\sqrt{2a^2 + c^2} \sqrt{13}}$.
Squaring both sides: $\frac{1}{2} = \frac{25a^2}{13(2a^2 + c^2)}$.
$13(2a^2 + c^2) = 50a^2 \implies 26a^2 + 13c^2 = 50a^2 \implies 13c^2 = 24a^2 \implies c^2 = \frac{24}{13}a^2 \implies c = \pm a \sqrt{\frac{24}{13}} = \pm a \frac{2\sqrt{6}}{\sqrt{13}}$.
Taking $a = \sqrt{13}$, we get $b = \sqrt{13}$ and $c = \pm 2\sqrt{6}$. Thus, the direction ratios are $\sqrt{13}, \sqrt{13}, 2\sqrt{6}$.

(C) Step $1$: Identify the normal vector $\vec{n}$ to the plane. The lines are parallel as they have the same direction vector $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.
Step $2$: Let $A = (0, 4, -1)$ and $B = (2, 1, 0)$ be points on the lines. The vector $\vec{AB} = (2-0)\hat{i} + (1-4)\hat{j} + (0-(-1))\hat{k} = 2\hat{i} - 3\hat{j} + \hat{k}$.
Step $3$: The normal vector $\vec{n} = \vec{AB} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(6-2) - \hat{j}(-4-1) + \hat{k}(4+3) = 4\hat{i} + 5\hat{j} + 7\hat{k}$.
Step $4$: The equation of the plane is $4(x-2) + 5(y-1) + 7(z-0) = 0$, which simplifies to $4x + 5y + 7z - 13 = 0$.
Step $5$: The distance $d$ from $(2, 5, 10)$ to the plane is $d = \frac{|4(2) + 5(5) + 7(10) - 13|}{\sqrt{4^2 + 5^2 + 7^2}} = \frac{|8 + 25 + 70 - 13|}{\sqrt{16 + 25 + 49}} = \frac{90}{\sqrt{90}} = \sqrt{90}$.
Step $6$: Therefore, $d^2 = 90$.

(A) Step $1$: Identify the position vector of a point on the plane. Let $\vec{a} = -\hat{i} + 2\hat{j}$.
Step $2$: Find the first direction vector $\vec{b}$ by the vector joining points $A(-1, 2, 0)$ and $B(2, 2, -1)$. $\vec{b} = (2 - (-1))\hat{i} + (2 - 2)\hat{j} + (-1 - 0)\hat{k} = 3\hat{i} - \hat{k}$.
Step $3$: Find the second direction vector $\vec{c}$ from the line $\frac{x - 1}{1} = \frac{y + 1/2}{1} = \frac{z + 1}{-1}$. The direction ratios are $(1, 1, -1)$, so $\vec{c} = \hat{i} + \hat{j} - \hat{k}$.
Step $4$: The parametric form of the plane is $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$.
Substituting the vectors: $\vec{r} = (-\hat{i} + 2\hat{j}) + \lambda(3\hat{i} - \hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$.
This matches option $A$.

(B) Step $1$: Convert the plane equation to intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Step $2$: Given $2x + 3y + z = 6$, divide by $6$ to get $\frac{2x}{6} + \frac{3y}{6} + \frac{z}{6} = 1$, which simplifies to $\frac{x}{3} + \frac{y}{2} + \frac{z}{6} = 1$.
Step $3$: The intercepts are $a = 3$, $b = 2$, and $c = 6$. Thus, the coordinates of $A, B, C$ are $(3, 0, 0)$, $(0, 2, 0)$, and $(0, 0, 6)$.
Step $4$: The volume of a tetrahedron with vertices at the origin and intercepts $a, b, c$ is given by $V = \frac{1}{6} |abc|$.
Step $5$: $V = \frac{1}{6} \times 3 \times 2 \times 6 = 6 \text{ cubic units}$.

(D) The given equation is $\vec{r} = \lambda(\hat{i} + 3\hat{k}) + \mu(\hat{i} + \hat{j} + 2\hat{k}) + 2\hat{j}$.
Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Then $x = \lambda + \mu$, $y = 2 + \mu$, $z = 3\lambda + 2\mu$.
From $y = 2 + \mu$, we get $\mu = y - 2$.
Substitute $\mu$ into $x = \lambda + \mu$: $x = \lambda + y - 2 \implies \lambda = x - y + 2$.
Substitute $\lambda$ and $\mu$ into $z = 3\lambda + 2\mu$: $z = 3(x - y + 2) + 2(y - 2) = 3x - 3y + 6 + 2y - 4 = 3x - y + 2$.
So, the equation of the plane is $3x - y - z = -2$, or $3x - y - z + 2 = 0$.
For the intercept form, $3x - y - z = -2 \implies \frac{x}{-2/3} + \frac{y}{2} + \frac{z}{2} = 1$.
Thus, $a = -2/3$, $b = 2$, $c = 2$.
Then $a + b + c = -2/3 + 2 + 2 = -2/3 + 4 = 10/3$.

(A) Step $1$: Find the normal vector $\vec{n_1}$ to the plane. The plane is parallel to $\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} - 2\hat{k}$.
Step $2$: $\vec{n_1} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -1 \\ 1 & -2 & -2 \end{vmatrix} = \hat{i}(-4 - 2) - \hat{j}(-6 + 1) + \hat{k}(-6 - 2) = -6\hat{i} + 5\hat{j} - 8\hat{k}$.
Step $3$: The normal to the $xy$-plane is $\vec{n_2} = \hat{k} = 0\hat{i} + 0\hat{j} + 1\hat{k}$.
Step $4$: The angle $\theta$ between the planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Step $5$: $\vec{n_1} \cdot \vec{n_2} = (-6)(0) + (5)(0) + (-8)(1) = -8$. So, $|\vec{n_1} \cdot \vec{n_2}| = 8$.
Step $6$: $|\vec{n_1}| = \sqrt{(-6)^2 + 5^2 + (-8)^2} = \sqrt{36 + 25 + 64} = \sqrt{125} = 5\sqrt{5}$. $|\vec{n_2}| = 1$.
Step $7$: $\cos \theta = \frac{8}{5\sqrt{5}}$. Thus, $\theta = \cos^{-1} \left( \frac{8}{5\sqrt{5}} \right)$.

(A) Step $1$: Since the plane is parallel to the $z$-axis, its normal vector $\vec{n}$ is perpendicular to the $z$-axis vector $\vec{k} = (0, 0, 1)$. Thus, the $z$-component of the normal vector is $0$. The equation of the plane is $3x - 2y - q = 0$, which matches this condition.
Step $2$: The plane passes through $A(1, p, 2)$. Substituting these coordinates into the equation: $3(1) - 2(p) - q = 0 \implies 3 - 2p - q = 0$.
Step $3$: The plane passes through $B(3, 2, 4)$. Substituting these coordinates into the equation: $3(3) - 2(2) - q = 0 \implies 9 - 4 - q = 0 \implies 5 - q = 0 \implies q = 5$.
Step $4$: Substitute $q = 5$ into the equation from Step $2$: $3 - 2p - 5 = 0 \implies -2p - 2 = 0 \implies p = -1$.
Step $5$: Thus, $p = -1$ and $q = 5$.

(C) The plane equation is $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$.
Setting $y=0, z=0$ gives $x=2$, so $A = (2, 0, 0)$.
Setting $x=0, z=0$ gives $y=3$, so $B = (0, 3, 0)$.
Setting $x=0, y=0$ gives $z=4$, so $C = (0, 0, 4)$.
The vectors $\vec{AB} = B - A = (-2, 3, 0)$ and $\vec{AC} = C - A = (-2, 0, 4)$.
The cross product $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ -2 & 0 & 4 \end{vmatrix} = \hat{i}(12 - 0) - \hat{j}(-8 - 0) + \hat{k}(0 - (-6)) = 12\hat{i} + 8\hat{j} + 6\hat{k}$.
The area of triangle $ABC$ is $\frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{12^2 + 8^2 + 6^2} = \frac{1}{2} \sqrt{144 + 64 + 36} = \frac{1}{2} \sqrt{244} = \frac{1}{2} \sqrt{4 \times 61} = \sqrt{61}$.

(D) The normal vector to the plane $\vec{r} \cdot (3\hat{i} - 2\hat{j} + 7\hat{k}) = 32$ is $\vec{n} = 3\hat{i} - 2\hat{j} + 7\hat{k}$.
The direction vector of the line is $\vec{v} = 5\hat{i} + b\hat{j} + 3\hat{k}$.
Since the line is parallel to the plane, the line is perpendicular to the normal vector of the plane. Therefore, the dot product of $\vec{n}$ and $\vec{v}$ must be zero:
$\vec{n} \cdot \vec{v} = 0$
$(3)(5) + (-2)(b) + (7)(3) = 0$
$15 - 2b + 21 = 0$
$36 - 2b = 0$
$2b = 36$
$b = 18$.