Find the shortest distance between the skew lines:

$\frac{x-3}{3} = \frac{8-y}{1} = \frac{z-3}{1}$

$\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$

Let $l, m, n$ be the direction cosines of the line of shortest distance. As it is perpendicular to given lines,
$3l - m + n = 0$
$-3l + 2m + 4n = 0$
$\implies \frac{l}{-6} = \frac{m}{-15} = \frac{n}{3}$
$\implies \frac{l}{-2} = \frac{m}{-5} = \frac{n}{1}$
$\implies$ distance $= \left |\frac{(3- (-3)).(-2) + (8-(-7)).(-5) + (3-6).1}{\sqrt{{(-2)}^2 + {(-5)}^2 + 1^2}} \right | = \left |\frac{-90}{\sqrt{30}} \right | = 3\sqrt{30}$