Let the plane $P$ pass through the intersection of the planes $2x + 3y - z = 2$ and $x + 2y + 3z = 6$,and be perpendicular to the plane $2x + y - z + 1 = 0$. If $d$ is the distance of $P$ from the point $(-7, 1, 1)$,then $d^2$ is equal to:

The distance of the point $(-1, 2, -2)$ from the line of intersection of the planes $2x + 3y + 2z = 0$ and $x - 2y + z = 0$ is:

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

The point of intersection of the line joining the points $\bar{i} + 2\bar{j} + \bar{k}$ and $2\bar{i} - \bar{j} - \bar{k}$ and the plane passing through the points $\bar{i}, 2\bar{j}, 3\bar{k}$ is:

The coordinates of the point where the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ meets the plane $x + y - z = 3$ are

$A$ plane $\pi_1$ contains the vectors $\bar{i}+\bar{j}$ and $\bar{i}+2\bar{j}$. Another plane $\pi_2$ contains the vectors $2\bar{i}-\bar{j}$ and $3\bar{i}+2\bar{k}$. $\bar{a}$ is a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If the angle $\theta$ between $\bar{a}$ and $\bar{i}-2\bar{j}+2\bar{k}$ is acute,then $\theta=$