A Tangent Through the Orthocenter

Let $\omega$ be the circumcircle of an acute triangle $ABC$ with orthocenter $H$. The tangent line at $H$ to the circumcircle of $\triangle HBC$ meets $\omega$ at points $X$ and $Y$. Given $HA=3$, $HX=2$, $HY=6$, the area of $\triangle ABC$ in simplest form is $m\sqrt n$, where $m$ and $n$ are positive integers and $n$ is squarefree. Find $m+n$.