= Solution
For nonparallel lines, $\mathbf n=\mathbf m_1\times\mathbf m_2$ is perpendicular to both direction vectors. Every displacement between the lines differs from $\mathbf a_2-\mathbf a_1$ by a vector in their span, so its component along $\mathbf n$ is fixed. Varying the two parameters cancels the component in that span. The <shortest distance between skew lines> is therefore
$$
\boxed{d=\frac{|(\mathbf a_2-\mathbf a_1)\cdot(\mathbf m_1\times\mathbf m_2)|}{|\mathbf m_1\times\mathbf m_2|}.}
$$
This also gives zero for intersecting nonparallel lines. For parallel lines the denominator vanishes and the correct separate formula is $| (\mathbf a_2-\mathbf a_1)\times\mathbf m_1|/|\mathbf m_1|$.
The given lines have vector forms
$$
\mathbf r_1=(2,3,-1)+t(2,1,2),\qquad
\mathbf r_2=(-2,-1,1)+u(1,2,1).
$$
Here $\mathbf m_1\times\mathbf m_2=(-3,0,3)$ and $\mathbf a_2-\mathbf a_1=(-4,-4,2)$. Their <scalar triple product> is $18$, giving
$$
\boxed{d=\frac{18}{3\sqrt2}=3\sqrt2.}
$$
As a constructive check, the closest parameters are $t=2/3$ and $u=7/3$, where the connecting vector is $(3,0,-3)$ and is perpendicular to both lines.
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