Let $EF$ meet $BC$ at point $S$, and let $\omega$ be tangent to $BC$ at point $D$. Because $(B, C; D, S) = -1$, we have $SD \cdot SM = SB \cdot SC = SP \cdot SQ$, that is, $D, M, P, Q$ are concyclic. Let $\odot(AEF)$ meet $\Omega$ again at point $T$, and let $AI$ meet $\Omega$ again at point $N$. Then, since $\triangle TBF \sim \triangle TCE$, we know $\frac{TB}{TC} = \frac{BF}{CE} = \frac{BD}{CE}$, that is, $T, D, N$ are collinear.

Suppose $\Gamma$ meets $\omega$ again at point $K$, and let the radical axis of $\odot(AEF)$ and $\Gamma$ be $\ell$. Considering $\odot(AEF)$, $\Gamma$, $\omega$, by the radical center theorem we know $\ell, DK, EF$ are concurrent; considering next $\odot(AEF), \Gamma, \Omega$, by the radical center theorem we know $\ell, PQ, AZ$ are concurrent, so $\ell, DK, EF, AZ$ are concurrent at point $X$, and $A, T, K, D$ are concyclic.

Therefore
$$\angle AKD = \angle ATD = \angle ATN = \angle(AN, BC) = \angle RMD = \angle RKD$$
that is, $R, A, K$ are collinear, which completes the proof.