The equation of the plane passing through the origin and containing the line $\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{5}$ is:

The distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$ is (in $\text{ units}$)

Let $P$ be the point of intersection of the line $\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}$ and the plane $x + y + z = 2$. If the distance of the point $P$ from the plane $3x - 4y + 12z = 32$ is $q$,then $q$ and $2q$ are the roots of the equation:

If the distance between the plane $ax - 2y + z = k$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$, then $|k|$ is

The line passing through $(1, 1, -1)$ and parallel to the vector $\hat{i} + 2 \hat{j} - \hat{k}$ meets the line $\frac{x - 3}{-1} = \frac{y + 2}{5} = \frac{z - 2}{-4}$ at $A$ and the plane $2 x - y + 2 z + 7 = 0$ at $B$. Then $AB = $

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,if