 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 $\sqrt{6}$. The planes through the two lines with this common normal have separationThis 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 $2\sqrt{6}$. Thus the shortest distance is $\boxed{2\sqrt{6}}$.