= Solution
A <normal vector> to the <plane> is $n=(3,4,12)^T$, with <Euclidean norm> $\|n\|=\sqrt{9+16+144}=13$. For $p=(1,1,1)^T$, $n\cdot p=19$, so the <distance from a point to a plane> is
$$
\boxed{d=\frac{|n\cdot p-26|}{\|n\|}=\frac7{13}.}
$$
To verify that this is the minimum, the displacement to the <closest point on a plane> is parallel to $n$. Put $q=p+tn$ and require $n\cdot q=26$. This gives $t=7/169$ and
$$
q=\frac1{169}(190,197,253)^T.
$$
For every other point $r$ on the <plane>, the displacement $r-q$ is orthogonal to $q-p$. The <Pythagorean theorem> gives $\|r-p\|^2=\|q-p\|^2+\|r-q\|^2\geq(7/13)^2$, proving minimality.
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