Plane Questions in English

(C) The equation of the plane is $6x - 3y + 2z = 6$. Dividing by $6$, we get $\frac{x}{1} + \frac{y}{-2} + \frac{z}{3} = 1$. Thus, the intercepts are $a = 1, b = -2, c = 3$.
The normal vector to the plane is $\vec{n} = 6\hat{i} - 3\hat{j} + 2\hat{k}$. The magnitude is $|\vec{n}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.
The direction cosines are $l = \frac{6}{7}, m = -\frac{3}{7}, n = \frac{2}{7}$.
The perpendicular distance $p$ from the origin to the plane $Ax + By + Cz + D = 0$ is $p = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. Here $p = \frac{|-6|}{7} = \frac{6}{7}$.
Now, calculate $|al + bm + cn| = |(1)(\frac{6}{7}) + (-2)(-\frac{3}{7}) + (3)(\frac{2}{7})| = |\frac{6}{7} + \frac{6}{7} + \frac{6}{7}| = |\frac{18}{7}|$.
Since $p = \frac{6}{7}$, we have $|al + bm + cn| = 3 \times \frac{6}{7} = 3p$.

(A) Let the intercepts of the plane on the coordinate axes be $a, b, c$. Thus, the points are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$.
The centroid of $\triangle ABC$ is given by $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.
Given that the centroid is $(1, r, r^2)$, we have:
$\frac{a}{3} = 1 \Rightarrow a = 3$
$\frac{b}{3} = r \Rightarrow b = 3r$
$\frac{c}{3} = r^2 \Rightarrow c = 3r^2$
The equation of the plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Substituting the values of $a, b, c$, we get $\frac{x}{3} + \frac{y}{3r} + \frac{z}{3r^2} = 1$.
Since the plane passes through $(5, 5, -12)$, we have:
$\frac{5}{3} + \frac{5}{3r} - \frac{12}{3r^2} = 1$
Multiplying by $3r^2$, we get $5r^2 + 5r - 12 = 3r^2$, which simplifies to $2r^2 + 5r - 12 = 0$.
Factoring the quadratic equation: $(2r - 3)(r + 4) = 0$.
Thus, $r = \frac{3}{2}$ or $r = -4$.

(A) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Substituting the point $(2, -1, -3)$, we get $a(x - 2) + b(y + 1) + c(z + 3) = 0$.
Since the plane is parallel to the lines with direction ratios $(3, 2, -4)$ and $(2, -3, 2)$, the normal vector $(a, b, c)$ must be perpendicular to these direction vectors.
Thus, $3a + 2b - 4c = 0$ and $2a - 3b + 2c = 0$.
Using the cross product to find the normal vector $(a, b, c) = (3, 2, -4) \times (2, -3, 2) = \begin{vmatrix} i & j & k \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{vmatrix} = i(4 - 12) - j(6 + 8) + k(-9 - 4) = -8i - 14j - 13k$.
Taking the normal vector as $(8, 14, 13)$, the equation of the plane is $8(x - 2) + 14(y + 1) + 13(z + 3) = 0$.
Expanding this, we get $8x - 16 + 14y + 14 + 13z + 39 = 0$, which simplifies to $8x + 14y + 13z + 37 = 0$.

(B) Given point $P(a, b, c)$.
Perpendicular $PA$ is drawn to the $YZ$-plane. The coordinates of $A$ are $(0, b, c)$.
Perpendicular $PB$ is drawn to the $ZX$-plane. The coordinates of $B$ are $(a, 0, c)$.
The origin $O$ is $(0, 0, 0)$.
The plane passes through $O(0, 0, 0)$, $A(0, b, c)$, and $B(a, 0, c)$.
The normal vector $\vec{n}$ to the plane is given by $\vec{OA} \times \vec{OB}$.
$\vec{OA} = 0\hat{i} + b\hat{j} + c\hat{k}$
$\vec{OB} = a\hat{i} + 0\hat{j} + c\hat{k}$
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & b & c \\ a & 0 & c \end{vmatrix} = \hat{i}(bc - 0) - \hat{j}(0 - ac) + \hat{k}(0 - ab) = bc\hat{i} + ac\hat{j} - ab\hat{k}$.
The equation of the plane passing through the origin is $bcx + acy - abz = 0$.

(A) Let the equation of the plane after rotation be $P_{3}: \ell x+my+nz=0$.
The line of intersection of the planes $P_{1}: \ell x+my=0$ and $P_{2}: z=0$ is the line where $\ell x+my=0$ and $z=0$.
The normal vectors are $\vec{n}_{1} = (\ell, m, 0)$ and $\vec{n}_{3} = (\ell, m, n)$.
The angle $\alpha$ between the planes $P_{1}$ and $P_{3}$ is given by $\cos \alpha = \frac{|\vec{n}_{1} \cdot \vec{n}_{3}|}{|\vec{n}_{1}| |\vec{n}_{3}|}$.
$\cos \alpha = \frac{|\ell^{2}+m^{2}|}{\sqrt{\ell^{2}+m^{2}} \sqrt{\ell^{2}+m^{2}+n^{2}}} = \sqrt{\frac{\ell^{2}+m^{2}}{\ell^{2}+m^{2}+n^{2}}}$.
Squaring both sides, $\cos^{2} \alpha = \frac{\ell^{2}+m^{2}}{\ell^{2}+m^{2}+n^{2}}$.
$\Rightarrow \cos^{2} \alpha (\ell^{2}+m^{2}+n^{2}) = \ell^{2}+m^{2}$.
$\Rightarrow n^{2} \cos^{2} \alpha = (\ell^{2}+m^{2})(1 - \cos^{2} \alpha) = (\ell^{2}+m^{2}) \sin^{2} \alpha$.
$\Rightarrow n^{2} = (\ell^{2}+m^{2}) \tan^{2} \alpha$.
$\Rightarrow n = \pm \sqrt{\ell^{2}+m^{2}} \tan \alpha$.
Substituting $n$ into the equation of $P_{3}$, we get $\ell x+my \pm z \sqrt{\ell^{2}+m^{2}} \tan \alpha = 0$.

(C) Let the points be $A(1, 2, 3)$ and $B(3, 4, 5)$.
The midpoint $M$ of the line segment $AB$ is given by $M = \left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{3+5}{2}\right) = (2, 3, 4)$.
The direction ratios of the line segment $AB$ are $(3-1, 4-2, 5-3) = (2, 2, 2)$.
Since the plane bisects $AB$ at right angles, the line $AB$ is normal to the plane. Thus, the normal vector is $\vec{n} = 2\hat{i} + 2\hat{j} + 2\hat{k}$, which can be simplified to $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
The equation of a plane passing through a point $\vec{a}$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.
Here, $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
Substituting these values, we get: $((x\hat{i} + y\hat{j} + z\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.
$(x-2)\hat{i} + (y-3)\hat{j} + (z-4)\hat{k} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.
$(x-2) + (y-3) + (z-4) = 0$.
$x + y + z - 9 = 0$, or $x + y + z = 9$.

(D) The plane passes through $(1, 2, -3)$ and $(2, -2, 1)$. The vector connecting these two points is $\vec{v} = (2-1)\hat{i} + (-2-2)\hat{j} + (1-(-3))\hat{k} = \hat{i} - 4\hat{j} + 4\hat{k}$.
Since the plane is parallel to the $X$-axis, its normal is perpendicular to the unit vector $\hat{i} = (1, 0, 0)$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{v}$ and $\hat{i}$:
$\vec{n} = \vec{v} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -4 & 4 \\ 1 & 0 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0-4) + \hat{k}(0-(-4)) = 4\hat{j} + 4\hat{k}$.
We can simplify the normal vector to $\vec{n}' = (0, 1, 1)$.
The equation of the plane passing through $(1, 2, -3)$ with normal $(0, 1, 1)$ is:
$0(x-1) + 1(y-2) + 1(z+3) = 0$
$y - 2 + z + 3 = 0$
$y + z + 1 = 0$.

(C) The equations of the given planes are $x+y+2z-6=0$ and $2x-y+z-9=0$.
Comparing these with the general form $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$, we get the normal vectors $\vec{n_1} = (1, 1, 2)$ and $\vec{n_2} = (2, -1, 1)$.
The angle $\theta$ between the two planes is given by the formula:
$\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$.
Substituting the values:
$\cos \theta = \frac{|(1)(2) + (1)(-1) + (2)(1)|}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+(-1)^2+1^2}}$.
$\cos \theta = \frac{|2 - 1 + 2|}{\sqrt{1+1+4} \sqrt{4+1+1}} = \frac{3}{\sqrt{6} \times \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$, we have $\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(A) The equation of the plane is given by $2x - y + 2z - 1 = 0$.
The normal vector $\vec{n}$ to this plane is given by the coefficients of $x, y,$ and $z$, which is $\vec{n} = 2\hat{i} - \hat{j} + 2\hat{k}$.
The direction vector of the $X$-axis is $\vec{a} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.
The angle $\theta$ between the normal vector $\vec{n}$ and the $X$-axis is given by the formula $\cos \theta = \frac{|\vec{n} \cdot \vec{a}|}{|\vec{n}| |\vec{a}|}$.
Calculating the dot product: $\vec{n} \cdot \vec{a} = (2)(1) + (-1)(0) + (2)(0) = 2$.
Calculating the magnitudes: $|\vec{n}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ and $|\vec{a}| = \sqrt{1^2 + 0^2 + 0^2} = 1$.
Therefore, $\cos \theta = \frac{2}{3 \times 1} = \frac{2}{3}$.
Thus, $\theta = \cos^{-1} \frac{2}{3}$.

(A) Let the points be $A(\vec{a}+\vec{b})$, $B(\vec{b}+\vec{c})$, and $C(\vec{c}+\vec{a})$.
The vectors lying on the plane are $\vec{AB} = B - A = (\vec{b}+\vec{c}) - (\vec{a}+\vec{b}) = \vec{c} - \vec{a}$ and $\vec{AC} = C - A = (\vec{c}+\vec{a}) - (\vec{a}+\vec{b}) = \vec{c} - \vec{b}$.
The normal vector $\vec{n}$ to the plane is $\vec{AB} \times \vec{AC} = (\vec{c} - \vec{a}) \times (\vec{c} - \vec{b}) = \vec{c} \times \vec{c} - \vec{c} \times \vec{b} - \vec{a} \times \vec{c} + \vec{a} \times \vec{b} = \vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b}$.
The equation of the plane is $\vec{r} \cdot \vec{n} = \vec{A} \cdot \vec{n}$.
$\vec{A} \cdot \vec{n} = (\vec{a}+\vec{b}) \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) = [\vec{a} \vec{b} \vec{c}] + [\vec{b} \vec{c} \vec{a}] = 2[\vec{a} \vec{b} \vec{c}]$.
Thus, the equation is $\vec{r} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = 2[\vec{a} \vec{b} \vec{c}]$.

(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 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 $\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 15$.

(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 $\sqrt{13}, \sqrt{13}, 2\sqrt{6}$.

(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 $1$: 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 $\frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{12^2 + 8^2 + 6^2} = \frac{1}{2} \sqrt{144 + 64 + 36} = \frac{1}{2} \sqrt{244} = \frac{1}{2} \sqrt{4 \times 61} = \sqrt{61}$.

(C) Step $1$: Find two vectors in the plane. Let $\vec{AB} = (2-1)\hat{i} + (-1+2)\hat{j} + (-3-1)\hat{k} = \hat{i} + \hat{j} - 4\hat{k}$ and $\vec{AC} = (0-1)\hat{i} + (1+2)\hat{j} + (5-1)\hat{k} = -\hat{i} + 3\hat{j} + 4\hat{k}$.
Step $2$: Find the normal vector $\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -4 \\ -1 & 3 & 4 \end{vmatrix} = \hat{i}(4+12) - \hat{j}(4-4) + \hat{k}(3+1) = 16\hat{i} + 0\hat{j} + 4\hat{k}$.
Step $3$: Simplify the normal vector to $\vec{n}' = 4\hat{i} + \hat{k}$.
Step $4$: The equation of the plane is $4(x-1) + 0(y+2) + 1(z-1) = 0$, which simplifies to $4x + z - 5 = 0$.
Step $5$: The perpendicular distance from $(0,0,0)$ to $Ax+By+Cz+D=0$ is $d = \frac{|D|}{\sqrt{A^2+B^2+C^2}} = \frac{|-5|}{\sqrt{4^2+0^2+1^2}} = \frac{5}{\sqrt{17}}$.

(B) Let the intercepts of the plane on the $x, y, z$ axes be $a, b, c$ respectively. The coordinates of the points are $A(a, 0, 0)$, $B(0, b, 0)$, and $C(0, 0, c)$.
The centroid of $\triangle ABC$ is given by $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.
Given the centroid is $(1, r, r^2)$, we have $\frac{a}{3} = 1 \implies a = 3$, $\frac{b}{3} = r \implies b = 3r$, and $\frac{c}{3} = r^2 \implies c = 3r^2$.
The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Substituting the values: $\frac{x}{3} + \frac{y}{3r} + \frac{z}{3r^2} = 1$.
Multiplying the entire equation by $3r^2$, we get $r^2x + ry + z = 3r^2$.

(A) Let the coordinates of points $A$, $B$, and $C$ be $(a, 0, 0)$, $(0, b, 0)$, and $(0, 0, c)$ respectively.
The centroid of triangle $ABC$ is $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.
Given the centroid is $(2, -\frac{2}{3}, \frac{1}{2})$, we have $\frac{a}{3} = 2 \implies a = 6$, $\frac{b}{3} = -\frac{2}{3} \implies b = -2$, and $\frac{c}{3} = \frac{1}{2} \implies c = \frac{3}{2}$.
The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, which becomes $\frac{x}{6} + \frac{y}{-2} + \frac{z}{3/2} = 1$, or $\frac{x}{6} - \frac{y}{2} + \frac{2z}{3} = 1$.
Multiplying by $6$, we get $x - 3y + 4z - 6 = 0$.
The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here, $d = \frac{|-6|}{\sqrt{1^2 + (-3)^2 + 4^2}} = \frac{6}{\sqrt{1 + 9 + 16}} = \frac{6}{\sqrt{26}}$.

(A) Step $1$: Distance of point $(1, 2, 3)$ from origin $(0, 0, 0)$ is $d_1 = \sqrt{(1-0)^2 + (2-0)^2 + (3-0)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
Step $2$: Distance of point $(1, 2, 3)$ from plane $2x - 3y + z + k = 0$ is $d_2 = \frac{|2(1) - 3(2) + 1(3) + k|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \frac{|2 - 6 + 3 + k|}{\sqrt{4 + 9 + 1}} = \frac{|k - 1|}{\sqrt{14}}$.
Step $3$: Given product $d_1 \times d_2 = 7$, so $\sqrt{14} \times \frac{|k - 1|}{\sqrt{14}} = 7$.
Step $4$: $|k - 1| = 7$, which implies $k - 1 = 7$ or $k - 1 = -7$.
Step $5$: Thus, $k = 8$ or $k = -6$. Given the options, $k = 8$ is the correct value.

(A) $1$. The point $P$ is $(1, 2, 3)$.
$2$. The foot of the perpendicular from $P(1, 2, 3)$ to the $YZ$-plane $(x=0)$ is $A(0, 2, 3)$.
$3$. The foot of the perpendicular from $P(1, 2, 3)$ to the $ZX$-plane $(y=0)$ is $B(1, 0, 3)$.
$4$. The plane passes through the origin $O(0, 0, 0)$, $A(0, 2, 3)$, and $B(1, 0, 3)$.
$5$. The equation of a plane passing through the origin is $ax + by + cz = 0$.
$6$. Substituting $A(0, 2, 3)$: $2b + 3c = 0 \implies b = -\frac{3}{2}c$.
$7$. Substituting $B(1, 0, 3)$: $a + 3c = 0 \implies a = -3c$.
$8$. Let $c = -2$, then $a = 6$ and $b = 3$.
$9$. The equation is $6x + 3y - 2z = 0$.

(A) $1$. The coordinates of point $P$ are $(a, b, c)$.
$2$. The perpendicular $PA$ is drawn to the $XY$ plane. The $XY$ plane has the equation $z = 0$. Thus, the coordinates of $A$ are $(a, b, 0)$.
$3$. The perpendicular $PB$ is drawn to the $ZX$ plane. The $ZX$ plane has the equation $y = 0$. Thus, the coordinates of $B$ are $(a, 0, c)$.
$4$. The origin $O$ is $(0, 0, 0)$.
$5$. The equation of a plane passing through $(0, 0, 0)$, $(a, b, 0)$, and $(a, 0, c)$ is given by the determinant form:
$\begin{vmatrix} x & y & z \\ a & b & 0 \\ a & 0 & c \end{vmatrix} = 0$
$6$. Expanding along the first row: $x(bc - 0) - y(ac - 0) + z(0 - ab) = 0$
$7$. $xbc - yac - zab = 0$
$8$. Dividing the entire equation by $abc$: $\frac{xbc}{abc} - \frac{yac}{abc} - \frac{zab}{abc} = 0$
$9$. $\frac{x}{a} - \frac{y}{b} - \frac{z}{c} = 0$.

(A) Step $1$: Find the midpoint $M$ of $AB$. $M = (\frac{4-1}{2}, \frac{5+2}{2}, \frac{-10+1}{2}) = (\frac{3}{2}, \frac{7}{2}, -\frac{9}{2})$.
Step $2$: The normal vector $\vec{n}$ to the plane is $\vec{AB} = (-1-4)\hat{i} + (2-5)\hat{j} + (1-(-10))\hat{k} = -5\hat{i} - 3\hat{j} + 11\hat{k}$.
Step $3$: The equation of the plane is $(\vec{r} - \vec{OM}) \cdot \vec{n} = 0$, where $\vec{OM} = \frac{3}{2}\hat{i} + \frac{7}{2}\hat{j} - \frac{9}{2}\hat{k}$.
Step $4$: $\vec{r} \cdot \vec{n} = \vec{OM} \cdot \vec{n} = (\frac{3}{2})(-5) + (\frac{7}{2})(-3) + (-\frac{9}{2})(11) = -\frac{15}{2} - \frac{21}{2} - \frac{99}{2} = -\frac{135}{2}$.
Step $5$: Thus, $\vec{r} \cdot (-5\hat{i} - 3\hat{j} + 11\hat{k}) = -\frac{135}{2}$, which rearranges to $\vec{r} \cdot (-5\hat{i} - 3\hat{j} + 11\hat{k}) + \frac{135}{2} = 0$.