 Solution

A normal vector to the plane is $n=(3,4,12{)}^T$, with Euclidean norm $\Vert n\Vert =\sqrt{9+16+144}=13$. For $p=(1,1,1{)}^T$, $n\cdot p=19$, so the distance from a point to a plane isTo 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 $t=7/169$ andFor every other point $r$ on the plane, the displacement $r-q$ is orthogonal to $q-p$. The Pythagorean theorem gives $\Vert r-p{\Vert }^2=\Vert q-p{\Vert }^2+\Vert r-q{\Vert }^2\ge (7/13{)}^2$, proving minimality.