Let $Q$ be the mirror image of the point $P(1, 0, 1)$ with respect to the plane $S: x + y + z = 5$. If a line $L$ passing through $(1, -1, -1)$,parallel to the line $PQ$,meets the plane $S$ at $R$,then $QR^{2}$ is equal to

The point $\bar{i}-2 \bar{j}$ lies on a line parallel to the vector $2 \bar{i}+\bar{k}$. The point $\bar{i}+2 \bar{j}$ lies on a plane parallel to the vectors $2 \bar{j}-\bar{k}$ and $\bar{i}+2 \bar{k}$. Find the point of intersection of the line and the plane.

The coordinates of the point,where the line through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $XZ$-plane,are

The equation of the plane which is parallel to the line $\frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7}$ and passes through the points $(0, 0, 0)$ and $(3, -1, 2)$ is

For what value of $c$ does the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersect the curve $xy = c^2, z = 0$?

Let the lines $\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does $NOT$ lie on $P$?