= 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>,
$$
\mathbf p_*=\frac{\mathbf p\cdot\mathbf d}{\mathbf d\cdot\mathbf d}\mathbf d
=\left(\frac5{17},0,\frac{20}{17}\right).
$$
The squared distance is $|\mathbf p|^2-(\mathbf p\cdot\mathbf d)^2/|\mathbf d|^2=3-25/17$, giving
$$
\boxed{d=\sqrt{\frac{26}{17}}.}
$$
Both plane normals are orthogonal to $\mathbf d$, and $(\mathbf p-\mathbf p_*)\cdot\mathbf d=0$, checking the geometric constraints.
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