Let the foot of the perpendicular from a point $P(1,2,-1)$ to the straight line $L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}$ be $N$. Let a line be drawn from $P$ parallel to the plane $x+y+2z=0$ which meets $L$ at point $Q$. If $\alpha$ is the acute angle between the lines $PN$ and $PQ$,then $\cos \alpha$ is equal to $.....$

The image of the point $(3, 2, 1)$ in the plane $2x - y + 3z = 7$ is

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 symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

The length of the projection of the line segment joining the points $(5, -1, 4)$ and $(4, -1, 3)$ on the plane $x + y + z = 7$ is:

$\overrightarrow{AB} = 3\hat{i} - \hat{j} + \hat{k}$ and $\overrightarrow{CD} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ are two vectors. The position vectors of the points $A$ and $C$ are $6\hat{i} + 7\hat{j} + 4\hat{k}$ and $-9\hat{j} + 2\hat{k}$ respectively. Find the position vectors of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\overrightarrow{PQ}$ is perpendicular to both $\overrightarrow{AB}$ and $\overrightarrow{CD}$.