Note that $PE$ and $QF$ intersect on the antipode $A'$ of $A$. Moreover, $\angle PA'B=\angle CEA'$ since $EC \parallel BA'$, and $\angle BCA'=90^\circ-\angle C=\angle FEC$. Now, perform the following angle-chase: $\angle PQF=\angle PQB+\angle BQA'=\angle PA'B+\angle BCA'=\angle CEA'+\angle FEC=\angle FEA',$ and so quadrilateral $PQFE$ is cyclic. Thus, $BC,EF,PQ$ concur on the radical center of circles $(PQEF), (EFCB)$ and $\omega$ (which is possibly a point at infinity).

We will prove that $PQFE$ is cyclic then it will be trivial by radical axis. Invert around $A$ with $r^2=AF*AC=AE*AB$. $P,Q$ will go to the points on $EF$ such that $\angle P’BA=\angle Q’CA=90$ then by easy angle chasing and we will get cycle.

Suppose that $AA'$ is diameter of $(ABC)$ and $AA'$ intersects $EF$ at $J$. Then $\overline{P, E, A'}$ and $\overline{Q, F, A'}$. So $\overline{A'F} \cdot \overline{A'Q} = \overline{A'J} \cdot \overline{A'A} = \overline{A'E} \cdot \overline{A'P}$ or $P, Q, E, F$ lie on a circle. From this, notice that $BC, EF, PQ$ are radical axis of $((ABC), (BCFE)); ((BCFE), (EFQP)); ((EFQP), (ABC));$ we have $BC, EF, PQ$ concur