The distance of the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ from the point $(3, 4, 5)$ is given by

$A$ vector parallel to the line of intersection of the planes $\bar{r} \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $\bar{r} \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{-5}$ in the plane $2x - y + z + 3 = 0$ is the line:

On which of the following lines does the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ lie?

$A$ 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 $Q$. The length of the line segment $PQ$ equals $\qquad$ units.

Find the coordinates of the foot of the perpendicular drawn from the point $P(-1, 1, 2)$ to the plane $2x - 3y + z - 11 = 0$.