The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

The square of the distance of the image of the point $A(6, 1, 5)$ in the line $\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}$ from the origin is:

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is $4$,then the sum of all possible values of $a$ is equal to :

The length of the perpendicular drawn from the point $(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ is

$A$ triangle $ABC$ is formed by vertices $A(1, -1, 0)$,$B(3, 5, 3)$,and $C(-11, -5, 6)$. The equation of the internal angle bisector of $\angle A$ is:

If the line joining the points $(k, 3, 4)$ and $(4, 7, 8)$ is parallel to the line joining the points $(-1, -2, 1)$ and $(1, 2, l)$,then $k + l =$