= Solution
Let $\mathbf d=\mathbf f-\mathbf e$. A point on the line has <position vector> $\mathbf r=\mathbf e+t\mathbf d$. By bilinearity and antisymmetry of the <cross product>,
$$
\boxed{\mathbf r\times\mathbf d=\mathbf e\times\mathbf d+t\mathbf d\times\mathbf d=\mathbf e\times\mathbf f}.
$$
For a point with <position vector> $\mathbf c$, the area of the <parallelogram> spanned by $\mathbf c-\mathbf e$ and $\mathbf d$ equals base length times perpendicular <Euclidean distance>. Hence the shortest <Euclidean distance> to the line is
$$
\boxed{d=\frac{|(\mathbf c-\mathbf e)\times(\mathbf f-\mathbf e)|}{|\mathbf f-\mathbf e|}}.
$$
For $\mathbf c=(0,0,1)$, these quantities are
$$
(\mathbf c-\mathbf e)\times(\mathbf f-\mathbf e)=\left(\frac14,\frac14,\frac14\right),\qquad
|\mathbf f-\mathbf e|=\frac1{\sqrt2}.
$$
Thus $\boxed{d=\sqrt6/4}$. The perpendicular meets $EF$ at $(1/4,1/4,1/2)$, whose displacement from $C$ has precisely this length.
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