If the lines $2x = ky = -z$ and $6x = -y = -4z$ are perpendicular to each other, then the value of $k$ is:

$ABC$ is a triangle with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $(\lambda^3 + \mu^3 + 5)$ is equal to:

The vector equation of a line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

The equation of the line of the shortest distance between the lines $\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$ and $\frac{x - 1}{0} = \frac{y + 1}{-2} = \frac{z}{1}$ is

If $d_1$ is the shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$,and $d_2$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$,then the value of $\frac{32 \sqrt{3} d_1}{d_2}$ is:

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