The line drawn from $(4, -1, 2)$ to the point $(-3, 2, 3)$ meets a plane at right angles at the point $(-10, 5, 4)$. Find the equation of the plane.

If the points $(1, 1, \mu)$ and $(-3, 0, 1)$ are equidistant from the plane $\vec{r} \cdot (3\hat{i} + 4\hat{j} - 12\hat{k}) + 13 = 0$, then the values of $\mu$ are

Let $\pi_1$ be the plane passing through the point $2\hat{i}-\hat{j}+\hat{k}$ and perpendicular to the vector $a\hat{i}+2\hat{j}-3\hat{k}$, and $\pi_2$ be the plane passing through the point $\hat{i}+2\hat{j}-\hat{k}$ and perpendicular to the vector $\hat{i}-2\hat{j}+\hat{k}$. If $\theta$ is the angle between the planes $\pi_1$ and $\pi_2$ and $\cos \theta = -\sqrt{\frac{3}{7}}$, then the integral value of $a$ is:

The Cartesian equation of a plane which passes through the points $A(2,2,2)$ and makes equal non-zero intercepts on the coordinate axes is

$A$ plane $\pi$ passes through the points $(5,1,2)$, $(3,-4,6)$, and $(7,0,-1)$. If $p$ is the perpendicular distance from the origin to the plane $\pi$ and $l, m, n$ are the direction cosines of a normal to the plane $\pi$, then $|3l+2m+5n|=$

If the mirror image of the point $(2, 4, 7)$ in the plane $3x - y + 4z = 2$ is $(a, b, c)$,then $2a + b + 2c$ is equal to