Alternative Solution by PSC. Let us write $\alpha, \beta, \gamma$ for the angles of $A B C$. Since $A D B C$ is cyclic, we have $\angle G D A=180^{\circ}-\angle B D A=\gamma$. Similarly, we have

$$\angle G A D=180^{\circ}-\angle D A E=\angle E B D=\angle B E C=\angle B A C=\alpha$$

where we have also used the fact that $\ell_{B}$ and $\ell_{C}$ are parallel.

Thus, the triangles $A B C$ and $A G D$ are similar. Analogously, $A E F$ is also similar to them.

Since $A D$ is a common chord of $\omega$ and $\omega_{1}$ then $A D$ is perpendicular to $O O_{1}$. Thus,

$$\angle O O_{1} A=\frac{1}{2} \angle D O_{1} A=\angle D G A=\beta$$

Similarly, we have $\angle O O_{2} A=\gamma$. Since $O_{1}, A, O_{2}$ are collinear (as in the first solution) we get that $O O_{1} O_{2}$ is also similar to $A B C$. Their circumcentres are $P$ and $O$ respectively, thus $\angle P O O_{1}=\angle O A B=90^{\circ}-\gamma$.

Since $O O_{1}$ is perpendicular to $A D$, letting $X$ be the point of intersection of $O O_{1}$ with $G D$, we get that $\angle D X O_{1}=90^{\circ}-\gamma$. Thus $O P$ is parallel to $\ell_{B}$ and therefore to $\ell_{C}$ as well.