 Solution

Use parametric equations for the two lines,Their cross product is $\mathbf{u}\times \mathbf{v}=(-3,0,3)\ne 0$. Projection of the separation of the base points onto this common normal gives the shortest distance between skew lines:This projection is a lower bound on every connecting segment because adding either line's direction leaves its normal component unchanged. It is attained: imposing perpendicularity of $\mathbf{p}+t\mathbf{u}-\mathbf{q}-s\mathbf{v}$ to both directions gives $t=2/3$ and $s=7/3$. The corresponding points are $(10/3,11/3,1/3)$ and $(1/3,11/3,10/3)$; their difference is $(3,0,-3)$, perpendicular to both lines and of length $3\sqrt{2}$.