Solution:

Note that $AEF \sim ABC$. Let the vertices of the triangle whose area we wish to compute be $P, Q, R$, opposite $A, E, F$ respectively. Since $H, O$ are isogonal conjugates, line $AH$ passes through the circumcenter of $AEF$, so $QR \parallel BC$.

Let $M$ be the midpoint of $BC$. We claim that $M = P$. This can be seen by angle chasing at $E, F$ to find that $\angle PFB = \angle ABC$, $\angle PEC = \angle ACB$, and noting that $M$ is the circumcenter of $BFEC$. So, the height from $P$ to $QR$ is the height from $A$ to $BC$, and thus if $K$ is the area of $ABC$, the area we want is $\frac{QR}{BC} K$.

Heron's formula gives $K = 84$, and similar triangles $QAF, MBF$ and $RAE, MCE$ give $QA = \frac{BC}{2} \frac{\tan B}{\tan A}$, $RA = \frac{BC}{2} \frac{\tan C}{\tan A}$, so that $\frac{QR}{BC} = \frac{\tan B + \tan C}{2 \tan A} = \frac{\tan B \tan C - 1}{2} = \frac{11}{10}$,
since the height from $A$ to $BC$ is $12$. So our answer is $\frac{462}{5}$.