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

The equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to each of the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

The coordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}$ are

The angle between the lines $\frac{x-1}{l}=\frac{y+1}{m}=\frac{z}{n}$ and $\frac{x+1}{m}=\frac{y-3}{n}=\frac{z-1}{l}$,where $l > m > n$ and $l, m, n$ are roots of the equation $x^3+x^2-4x-4=0$,is

The angle between the lines $3x = 2y = -z$ and $-x = 6y = -4z$ is

Find the distance of a point $(2, 4, -1)$ from the line $\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9}$.