(By Zuming Feng) By Miquel's theorem (which can be shown by simple angle chasing), the circumcircles of triangles $AQR$, $BRP$, and $CPQ$ meet at a common point $X$. The key observation is that $XR \perp AB$, $XP \perp BC$, and $XQ \perp CA$. Indeed, if $P_1Q_1R_1$ is an inscribed equilateral triangle, and the circumcircles of triangles $AQ_1R_1$, $BR_1P_1$, and $CP_1Q_1$ meet at a common point $X$. Let $P$, $Q$, and $R$ be the feet of the perpendiculars from $X$ to the sides of the triangle. Quick angle chasing ($\angle RR_1X = \angle PP_1X = QQ_1X$) shows that right triangles $XPP_1$, $XQQ_1$, and $XRR_1$ are similar, and so triangles $PQR$ and $P_1Q_1R_1$ are similar. Clearly, $PQR$ is a smaller triangle, and this establishes our observation.

It is then straightforward to check that the perpendiculars from $A$ to $QR$, $B$ to $RP$, and $C$ to $PQ$ meet at the isogonal conjugate of $X$.