The coordinates of the foot of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ are:

Prove that the lines $x=p y+q, z=r y+s$ and $x=p^{\prime} y+q^{\prime}, z=r^{\prime} y+s^{\prime}$ are perpendicular if $p p^{\prime}+r r^{\prime}+1=0$.

The distance of the point $P(7, 10, 11)$ from the line $\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3}$ along the line $\frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6}$ 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:

If the coordinates of the points $A, B, C, D$ are $(1, 2, 3), (4, 5, 7), (-4, 3, -6)$ and $(2, 9, 2)$ respectively,then the angle between the lines $AB$ and $CD$ is

If the Cartesian equation of the line is $x-1=2y+3=3-z$,then its vector equation is