 Solution

Taking cross products of the two independent spanning vectors gives plane normal vectors proportional to $(-4,1,1)$ and $(4,-6,-1)$. Their cross product is $(5,0,20)$, so the intersection is the line $s\mathbf{d}$, with $\mathbf{d}=(1,0,4)$. The closest point to $\mathbf{p}=(1,1,1)$ is its orthogonal projection,The squared distance is $|\mathbf{p}{|}^2-(\mathbf{p}\cdot \mathbf{d}{)}^2/|\mathbf{d}{|}^2=3-25/17$, givingBoth plane normals are orthogonal to $\mathbf{d}$, and $(\mathbf{p}-{\mathbf{p}}_{\ast })\cdot \mathbf{d}=0$, checking the geometric constraints.