The line passing through $(1, 1, -1)$ and parallel to the vector $\hat{i} + 2 \hat{j} - \hat{k}$ meets the line $\frac{x - 3}{-1} = \frac{y + 2}{5} = \frac{z - 2}{-4}$ at $A$ and the plane $2 x - y + 2 z + 7 = 0$ at $B$. Then $AB = $

The point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane $3x+4y+5z=10$ is

If the line joining the points $A(1,0,0)$ and $B(0,0,1)$ is a normal to the plane $\pi$ which passes through the point $A$, then the angle between the planes $\pi$ and $x+y+z=6$ is

The equation of the locus of a point $P(x, y, z)$ such that its distance from the $X$-axis is equal to its distance from the plane $x+z=1$ is

Consider the line $L$ given by the equation $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$. Let $Q$ be the mirror image of the point $P_0(2,3,-1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$,and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to each of the planes $3x - 2y + z = 8$ and $x + y + z = 6$ is $lx + my + nz = 1$,then $4m + 2n - 31 =$