Htykuut

I am stuck on this problem -

Let $O$ and $H$ be the circumcenter and orthocenter of triangle $ABC$, respectively. Let $a$, $b$, and $c$ denote the side lengths.

Find $AH^2 + BH^2+CH^2$ if the circumradius is equal to $7$ and $a^2 + b^2 + c^2 = 432$.

So far, I've tried using vectors to tease out the solution, but nothing's working. I have found that $OH$ is $3$ but I'm not sure how that helps.

Consider the triangle obtained by drawing at each vertex of $ABC$ a line parallel to its opposite side. These three new lines form a new triangle $A'B'C'$ that can be seen to be similar to the original triangle, with doubled sides. Also, notice that the circumcenter of $A'B'C'$ is the orthocenter of $ABC$, which we denoted $H$. Thus, by Pythagoras,

begin{gather*}
(HC')^2 = (AC')^2 + AH^2 = a^2 + AH^2\
(HA')^2 = (BA')^2 + BH^2 = b^2 + BH^2\
(HB')^2 = (CB')^2 + CH^2 = c^2 + CH^2.
end{gather*}
Summing the three expressions above and using $HC' = HB' = HA' = 2 cdot 7 = 14$ (by the similarity relation), we have
$$
3 cdot 14^2 = a^2 + b^2 + c^2 + (CH^2 + AH^2 + BH^2) = 432 + (CH^2 + AH^2 + BH^2)
$$
implying that
$$
CH^2 + AH^2 + BH^2 = 588 - 432 = 156.
$$