The distance of the point $P(-3, 2, 3)$ from the line passing through $A(4, 6, -2)$ with direction ratios $\langle -1, 2, 3 \rangle$ is . . . . . . units.

The angle between the lines $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$ and $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ is

The direction cosines of a line which is perpendicular to the lines $\frac{x - 7}{2} = \frac{y + 17}{-3} = \frac{z - 6}{1}$ and $\frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 6}{-2}$ are...

If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ have a point in common,then $k=$

Let $O$ be the origin,and $M$ and $N$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\overrightarrow{OM} \cdot \overrightarrow{ON}$ is equal to:

Let the shortest distance between the lines $L : \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}, \lambda \geq 0$ and $L_1 : x+1 = y-1 = 4-z$ be $2\sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$,then which of the following is $NOT$ possible?