= Solution
Take points $p=(3,0,4)$ and $q=(-2,3,3)$ with directions $u=(1,3,-1)$ and $v=(0,1,-1)$. A common normal is $u\times v=(-2,1,1)$, of length $\sqrt6$. The <planes> through the two lines with this common normal have separation
$$
\frac{|(q-p)\cdot(u\times v)|}{|u\times v|}
=\frac{12}{\sqrt6}=2\sqrt6.
$$
This is a lower bound for the <distance between skew lines>. It is attained at $s=-1$ and $t=-4$: the two points are $(2,-3,5)$ and $(-2,-1,7)$, whose difference $(4,-2,-2)$ is perpendicular to both directions and has length $2\sqrt6$. Thus \b[the shortest distance is] $\boxed{2\sqrt6}$.
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