Solution
= Solution
Use the <normal vector> $\mathbf N=(-8,-3,-9)$ found from the edge <vectors>, rather than the <normal vector> to the origin-based <linear span> in part (b). The required <plane> satisfies $\mathbf N\cdot(\mathbf r-\mathbf a)=0$. Since $\mathbf N\cdot\mathbf a=5$,
$$
\boxed{-8x-3y-9z=5.}
$$
After dividing by $|\mathbf N|=\sqrt{154}$, its equation has a <unit normal>. Its perpendicular distance from the origin is thus $\boxed{5/\sqrt{154}}$.