Answer:
3600
Solution: Let the centers of $\omega_{1}$ and $\omega_{2}$ be $O_{1}$ and $O_{2}$. Let $D E$ intersect $\omega_{1}$ again at $Q$, and let $D F$ intersect $\omega_{2}$ again at $R$. Note that since $\omega_{1}$ and $\omega_{2}$ must be tangent to $A D$ at the same point (by equal tangents), so $A D$ must be the radical axis of $\omega_{1}$ and $\omega_{2}$, so $R Q E F$ is cyclic. Thus, we have
$$
\angle O_{1} Q R=\angle E Q R-\angle O_{1} Q E=180^{\circ}-\angle E F D-\angle O_{1} E Q=90^{\circ}
$$

Thus, we have $Q R$ is tangent to $\omega_{1}$, and similarly it must be tangent to $\omega_{2}$ as well.
Now, note that by Monge's theorem on $\omega, \omega_{1}$, and $\omega_{2}$, we have that $P$ must be the intersection of the external tangents of $\omega_{1}$ and $\omega_{2}$. Since $R Q$ is an external tangent, we have $P, Q$, and $R$ are collinear. Thus, by power of a point, we have $P E \cdot P F=P R \cdot P Q$. Note that $P R=10 \sqrt{30}$ and $P Q=12 \sqrt{30}$. Thus, we have $P E \cdot P F=3600$.
26. [16] Let triangle $A B C$ have incircle $\omega$, which touches $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Then, let $\omega_{1}$ and $\omega_{2}$ be circles tangent to $A D$ and internally tangent to $\omega$ at $E$ and $F$, respectively. Let $P$ be the intersection of line $E F$ and the line passing through the centers of $\omega_{1}$ and $\omega_{2}$. If $\omega_{1}$ and $\omega_{2}$ have radii 5 and 6 , respectively, compute $P E \cdot P F$.