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

Find the distance of the point $(-1, -2, -1)$ from the plane passing through the point $(1, 1, 1)$ and perpendicular to both lines $L_1: \frac{x-1}{1} = \frac{y-1}{0} = \frac{z-1}{-1}$ and $L_2: \frac{x-1}{0} = \frac{y-1}{1} = \frac{z-1}{-1}$.

What is the length of the perpendicular drawn from the origin to the plane passing through point $a$ and containing the line $r = b + \lambda c$?

The equation of the line passing through the intersection of the plane $x+2y+3z=4$ and the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}$ and parallel to the vector $(2\hat{i}-3\hat{j}) \times (\hat{i}+2\hat{j}-\hat{k})$ is

The equation of the plane in which the lines $\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{-5}$ and $\frac{x - 8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}$ lie,is

Angle between the lines of intersection of the planes $x-y=0, 2x+y+z=0$ and $2x-z=0, x+y-3z=0$ is (in $^{\circ}$)