[asy][asy] size(7cm); defaultpen(fontsize(10pt)); pair A = dir(120), B =dir(210), C = dir(330), O = origin, P = (A+2B)/3, Q = intersectionpoint(circumcircle(A,P,O),C--C+dir(C--A)*abs(C-A)*0.99), H = orthocenter(O,P,Q); draw(A--B--C--A, heavycyan); draw(circumcircle(A,P,O), blue); draw(circumcircle(B,P,O)^^circumcircle(C,Q,O), orange); draw(P--O--Q--P, blue); draw(A--O--B, heavygreen); draw(O--H, blue); dot("$A$", A, dir(110)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$O$", O, dir(0)); dot("$P$", P, dir(180)); dot("$Q$", Q, dir(0)); dot("$H$", H, dir(280)); [/asy][/asy] Switch to $A$-indexing, and let $H$ be the orthocenter of $OPQ$. Then we have \[\measuredangle PHO = \measuredangle OQP = \measuredangle OAP = \measuredangle ABO = \measuredangle PBO,\]so $H$ lies on $(BPO)$. Similarly $H$ also lies on $(CQO)$, so it must lie on $\overline{BC}$ by Miquel point.

Let $M$ and $N$ be the midpoints of $AB$ and $BC$, respectively. Let's say the perpendicular from $O$ to $PQ$ intersects to $AC$ and $PQ$ at $H$ and $T$, respectively. Let $R = OP \cap HQ$. By Simson's theorem $M$, $T$ and $N$ are collinear. Notice that \[\angle OPC = \angle OPN = \angle OTN = \angle HTN = \angle AHO\](since $MN \parallel AC$) then $OHCP$ is cyclic. Analogously $OHAQ$ is cyclic. We see that \[\angle ACB = \angle TNP = \angle TOP = \angle ROH\]but \[\angle OHR = \angle OHQ = \angle OAQ = 90^{\circ} - \angle ACB = 90^{\circ} - \angle ROH\]then $PO \perp QH$. Therefore $H$ is orthocenter of $\triangle OPQ$, which lies on $AC$ $\blacksquare$

oo cute Define $R$ as the $(AQO)\cap AC.$ We claim that $R$ is the orthocenter of $OPQ$. For $RO\perp PQ$, note that $\angle OQP=90-A$ and $A=\angle QAR=180-\angle QOR$. So $RO\perp PQ$. For $RP\perp QO$, note that $\angle ORP=\angle OCP=90-A$ and $$\angle QPR=\angle QPO+\angle RPQ=\angle OBA+\angle OCA=90-C+90-B\implies RP\perp QO.$$