If the lines $\frac{x - 1}{c} = \frac{y + 2}{-2} = \frac{z - 3}{4}$ and $\frac{x - 5}{1} = \frac{y - 3}{1} = \frac{z + 1}{c}$ are parallel,then $c = ....$

Find the foot of the perpendicular drawn from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$.

An angle between the lines whose direction cosines are given by the equations $l + 3m + 5n = 0$ and $5lm - 2mn + 6nl = 0$ is

The distance of the point $P(4, 6, -2)$ from the line passing through the point $(-3, 2, 3)$ and parallel to a line with direction ratios $3, 3, -1$ is equal to:

Let $A$ and $B$ be two distinct points on the line $L : \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the line $L$. If $O$ is the origin,then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:

The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is: