Find the angle between the lines $\frac{x - 2}{3} = \frac{y + 1}{-2}; z = 2$ and $\frac{x - 1}{1} = \frac{2y + 3}{3} = \frac{z + 5}{2}$.

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:

The equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$ is:

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

What is the point of intersection of the line joining the points $(3, 4, 1)$ and $(5, 1, 6)$ with the $xy$-plane?

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then what is the value of $k$?