= Solution
The plane has <normal vector> $\mathbf n=(2,3,6)$ of length seven. Projecting the displacement from any point on the plane onto the <unit normal> gives the <distance from a point to a plane>
$$
\boxed{d=\frac{|2x_0+3y_0+6z_0-30|}{7}.}
$$
For example the perpendicular foot is $\mathbf r_0-\mathbf n(\mathbf n\cdot\mathbf r_0-30)/49$. Substitution places it on the plane, and its displacement from $\mathbf r_0$ is parallel to $\mathbf n$, verifying that this is the minimum distance.
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