= Solution
From the line point $r_0=(1,-1,0)$ to $P=(2,2,1)$, take $w=P-r_0=(1,3,1)$ and line direction $d=(2,3,1)$. <Orthogonal projection> gives the nearest parameter $t=(w\cdot d)/(d\cdot d)=12/14=6/7$. The perpendicular displacement has squared <Euclidean norm>
$$
|w|^2-\frac{(w\cdot d)^2}{|d|^2}=11-\frac{144}{14}=\frac57.
$$
Hence the <distance from a point to a line in three dimensions> is
$$
\boxed{\sqrt{5/7}.}
$$
The nearest point is $(19/7,11/7,6/7)$; subtracting it from $P$ gives a vector orthogonal to $d$, verifying the minimum.
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