The equation of the plane passing through the intersection of the planes $2x - 5y + z = 3$ and $x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is $x + 3y + 6z = k$,where $k$ is:

$A$ line with positive direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2x + y + z = 9$ at point $Q,$ then the length $PQ$ equals

If the equation of the plane passing through the points $(2,1,2)$ and $(1,2,1)$ and perpendicular to the plane $2x - y + 2z = 1$ is $ax + by + cz + d = 0$,then $\frac{a+b}{c+d} = $

Find the coordinates of the point of intersection of the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ and the plane $x + y - z = 3$.

If the point of intersection of the lines $r = \hat{i} - 6\hat{j} + (p \sec \alpha) \hat{k} + t(\hat{i} + 2\hat{j} + \hat{k})$ and $r = 4\hat{j} + \hat{k} + \lambda(2\hat{i} + (p \tan \alpha) \hat{j} + 2\hat{k})$ is $8\hat{i} + 8\hat{j} + 9\hat{k}$,(where $0 < \alpha < \frac{\pi}{2}$),then $p =$

The foot of the perpendicular drawn from the point $(1, 1, 1)$ to the plane $\pi_1$ is $(1, 3, 5)$. If $(2, 2, -1), (3, 4, 2), (3, 3, 0)$ are three points on the plane $\pi_2$,then the angle between the planes $\pi_1$ and $\pi_2$ is