Question: Prove that for the cone $$ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$$ having three mutually perpendicular generators, $a+b+c=0$.
Proceed: If the cone has three mutually perpendicular generators, then we have, generally $Lyz+Mzx+Nxy=0$, where $L,M,N$ are constants. How can I show that $a+b+c=0$?
In general, given a symmetric matrix $S$ and an invertible matrix $Q,$ we do not know that much about $Q^T SQ$ other than Sylvester's Law of Inertia. If the columns of $Q$ are "generators" then the diagonal of $Q^TSQ$ has all zero elements.
In your special case, symmetric matrix $H,$ we can take your three generators, normalize each to length one, make a square matrix $P$ with those columns, so that now $P^TP = PP^T = I.$ We know that $P^T H P$ has all diagonal elements zero. The thing that has changed is that the trace of $P^T HP$ is the same as the trace of $H,$ namely $a+b+c,$ which is therefore zero.