Notice that $A_{2}$ is the point of tangency of the excircle opposite $A$ to $B C$. Therefore, by considering the homothety centered at $A$ taking the excircle to the incircle, we notice that $A_{3}$ is the intersection of $\omega$ and the tangent line parallel to $B C$. It follows that $A_{1} B_{1} C_{1}$ is congruent to $A_{3} B_{3} C_{3}$ by reflecting through the center of $\omega$. We therefore need only find \left[A_{1} B_{1} C_{1}\right] /[A B C]. Since $$\frac{\left[A_{1} B C_{1}\right]}{[A B C]}=\frac{A_{1} B \cdot B C_{1}}{A B \cdot B C}=\frac{((9+10-13) / 2)^{2}}{9 \cdot 10}=\frac{1}{10}$$ and likewise \left[A_{1} B_{1} C\right] /[A B C]=49 / 130 and \left[A B_{1} C_{1}\right] /[A B C]=4 / 13, we get that $$\frac{\left[A_{3} B_{3} C_{3}\right]}{[A B C]}=1-\frac{1}{10}-\frac{49}{130}-\frac{4}{13}=\frac{14}{65}$$