$OABCD$ tetrahedron with $OA ⊥ OB ⊥ OC ⊥ OA$

Question

I've got stuck at this problem:

Let $OABCD$ be a tetrahedron with $OA ⊥ OB ⊥ OC ⊥ OA$. If $OH$ is the orthocentre of triangle $ABC$, show that $OH$ is perpendicular on plan $(ABC)$. Then prove that $OA^2 + OB^2 + OC^2 = 9OH^2$.

Since I'm not familiar to any of these programs where you can make geometry drawings, I'm just showing my pencil drawing below (please let me know if the drawing is unclear):

The first thing that came to my mind was Theorem of Three Perpendiculars, but I wasn't able to solve it.

I would really apreciate some hints.

Thanks!