The image of the point with position vector $(\hat{i}+3 \hat{j}+4 \hat{k})$ in the plane $r \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$ is

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is

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 $p =$

$A$ 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

The plane containing the point $(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is

Let two planes be given by $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Based on the given information,find the equation of the plane passing through the intersection of $P_1$ and $P_2$ and the point $(3, 2, 1)$.