Throughout the solution we use some known properties of perspective-orthologic triangles. At first, line $PQ$ passes through the second center $R$ of orthology of $ABC,A_1B_1C_1.$ Define point $A_2$ as symmetric to $A_1$ over perpendicular bisector of $BC,$ and $B_2,C_2$ similarly. We see (arcs are oriented) $$\angle (B_1C_1;AB)=\widehat{B_1A} +\widehat{C_1B} =\widehat{CB_2} +\widehat{AC_2} =\angle (AC;B_2C_2),$$therefore perpendicular from $A$ to $B_2C_2$ isogonal to $AR$ in angle $BAC;$ also lines $AA_1,AA_2$ isogonals in this angle. Now we see that $Q',R',$ which are isogonal conjugates of $Q,R$ in $ABC,$ are respectively centers of perspective and orthology of this triangle and $A_2B_2C_2.$ By the property,conic $(ABCQ'R')$ is equilateral hyperbola, and so passes through orthocenter of $ABC.$ But this conic is isogonal conjugate of $QR$ wrt $ABC.$ Thus the line passes through circumcenter of triangle.

Dang, I've gotta learn what perspective-orthologic triangles are. Anyways, here is my approach: Let $O$ be the circumcenter of $\triangle ABC$. We will prove that all lines $PQ$ go through point $O$, therefore all such lines concur. $\textbf{Lemma}$ $\angle QAC+\angle QCB+\angle QBA=90^{\circ}$ $\textit{Proof:}$ We will use the trig Ceva. Let $\angle QAB=t$, $\angle QAC=z$, $\angle QCA=x$, $\angle QCB=y$, $\angle QBC=\beta$, $\angle QBA=\alpha$. Using trig Ceva for the lines $AQ,BQ,CQ$ we have that: $$\frac{\sin t}{\sin z}\cdot\frac{\sin x}{\sin y}\cdot \frac{\sin \beta}{\sin\alpha}=1$$Let $\alpha+y+z=90^{\circ}+\varepsilon$, where $-90^{\circ}<\varepsilon<90^{\circ}$. Notice that: $$\angle AC_{1}B_{1}=\angle ABB_{1}=\angle ACB_{1}=\alpha$$$$\angle CA_{1}B_{1}=\angle CBB_{1}=\angle CAB_{1}=\beta$$$$\angle C_{1}A_{1}B=\angle C_{1}CB=\angle C_{1}AB=y$$$$\angle AB_{1}C_{1}=\angle ACC_{1}=\angle ABC_{1}=x$$$$\angle CBA_{1}=\angle CAA_{1}=\angle CB_{1}A_{1}=z$$$$\angle A_{1}C_{1}B=\angle A_{1}AB=\angle BCA_{1}=t$$Now, since we know that $PC_{1}\perp AB$, $PA_{1}\perp BC$, $PB_{1}\perp AC$ and using the angles we mentioned we have that $\angle B_{1}C_{1}P=90^{\circ}-\angle B_{1}C_{1}A-\angle C_{1}AB=90^{\circ}-\alpha-y$. Analogically, we have that $\angle PC_{1}A_{1}=90^{\circ}=90^{\circ}-t-x$, $\angle PA_{1}C_{1}=90^{\circ}-y-z$, $\angle PA_{1}B_{1}=90^{\circ}-t-\beta$, $\angle PB_{1}A_{1}=90^{\circ}-z-\alpha$, $\angle PB_{1}C_{1}=90^{\circ}-\beta-x$. Now, by using trig Ceva for $\triangle A_{1}B_{1}C_{1}$ and the intersecting lines $A_{1}P,B_{1}P,C_{1}P$ we have that: $$\frac{\sin\angle B_{1}C_{1}P}{\sin\angle PC_{1}A_{1}}\cdot\frac{\sin\angle C_{1}A_{1}P}{\sin\angle PA_{1}B_{1}}\cdot\frac{\sin\angle A_{1}B_{1}P}{\sin\angle PB_{1}C_{1}}=1$$$$\Longrightarrow \frac{\sin\left(90^{\circ}-\alpha-y\right)}{\sin\left(90^{\circ}-x-t\right)}\cdot\frac{\sin\left(90^{\circ}-y-z\right)}{\sin\left(90^{\circ}-t-\beta\right)}\cdot\frac{\sin\left(90^{\circ}-z-\alpha\right)}{\sin\left(90^{\circ}-\beta-x\right)}=1$$$$\Longrightarrow \frac{\sin (z-\varepsilon)}{\sin (t+\varepsilon)}\cdot\frac{\sin (y-\varepsilon)}{\sin (x+\varepsilon)}\cdot \frac{\sin(\alpha-\varepsilon)}{\sin (\beta+\varepsilon)}=1$$Since $-90^{\circ}<\varepsilon<90^{\circ}$ that means that $\sin\beta\sin x\sin t<\sin(t+\varepsilon)\sin(x+\varepsilon)\sin(\beta+\varepsilon)$ if $\varepsilon>0$ and with the opposite sign inequality for $\varepsilon<0$. Analogical inequality holds for $\alpha, y,z$. Therefore both equalities occur only if $\varepsilon=0\Longrightarrow \alpha+z+y=\beta+x+t=90^{\circ}$ and we've proven the Lemma. (Another way of proving this result is constructing the isogonal conjugate of point $Q$ for $\triangle A_{1}B_{1}C_{1}$, say $Q'$. Now we have that if $\varepsilon\neq 0$ the lines $C_{1}P$, $B_{1}P$, $A_{1}P$ would obviously not intersect, see the diagram for a visual representation). The Lemma implies that $\angle B_{1}C_{1}P=z$, $\angle PC_{1}A_{1}=\beta$, $\angle PA_{1}C_{1}=\alpha$, $\angle PA_{1}B_{1}=x$, $\angle PB_{1}A_{1}=y$, $\angle PB_{1}C_{1}=t$, therefore: $$\triangle B_{1}PA_{1}\sim\triangle AC_{1}B,\quad \triangle C_{1}PA_{1}\sim\triangle AB_{1}C,\quad \triangle B_{1}PC_{1}\sim\triangle CA_{1}B$$Also, we have that $\angle B_{1}C_{1}P=z=\angle QC_{1}A_{1}$, so $P$ and $Q$ are isogonal points wrt $\angle B_{1}C_{1}A_{1}$ and simmilarly $P,Q$ are isogonal wrt $\angle C_{1}A_{1}B_{1}$ and $\angle C_{1}B_{1}A_{1}$, so $P$ and $Q$ are isogonal conjugates for $\triangle A_{1}B_{1}C_{1}$. The problem can be finished with trilinear coordinates, here are Link1 and Link2 for reference. We know that: $$O=(\cos(\alpha+x):\cos(y+t):\cos(\beta+z))$$We can calculate $Q$, because $\text{dist}(Q,C_{1}B_{1})=\cos\beta\cdot C_{1}Q$ and $\text{dist}(Q,A_{1}C_{1})=\cos z\cdot A_{1}Q$, so wehave that $\frac{\text{dist}(Q,B_{1}C_{1})}{\text{dist}(Q,A_{1}C_{1})}=\frac{\cos\beta}{\cos z}$ and similarly $\frac{\text{dist}(Q,A_{1}C_{1})}{\text{dist}(Q,A_{1}B_{1})}=\frac{\cos\alpha}{\cos x}$, so: $$Q=(\text{dist}(Q,B_{1}C_{1}):\text{dist}(Q,A_{1}C_{1}):\text{dist}(Q,A_{1}B_{1})=(\cos\alpha\cos\beta:\cos \alpha\cos z: \cos x\cos z)$$Also, (from the second link) we know that if $Q=(X:Y:Z)$, then $P=(\frac{1}{X}:\frac{1}{Y}:\frac{1}{Z})$, so we have that: $$P=(\cos z\cos x:\cos x\cos \beta:\cos \alpha\cos \beta)$$Now we can use the collinearity formula for trilinear coordianates to finish the problem: $$O\in PQ\iff \begin{vmatrix} \cos(\alpha+x) & \cos(y+t) & \cos(\beta+z) \\ \cos \alpha\cos\beta & \cos \alpha\cos z & \cos z\cos x \\ \cos z\cos x & \cos x\cos \beta & \cos \alpha\cos \beta \\ \end{vmatrix}=0$$This should be relatively easy, here are some ideas: write out the determinant, rewrite $\cos(y+t)=-\cos((x+\alpha)+(\beta+z))=-\cos (x+\alpha)\cos (\beta+z)+\sin (x+\alpha)\sin (\beta+z)$, use $\cos(x+\alpha)=\cos x\cos\alpha-\sin x\sin\alpha$ and $\cos(\beta+z)=\cos\beta\cos z-\sin\beta \sin z$ and also the result from the Lemma to find a relationship for $\frac{\text{dist}(Q,B_{1}C_{1})}{\text{dist}(Q,A_{1}B_{1})}$ and things should cancel out, but I didn't manage to finish it before the end of the olympiad. Could someone please look over this and maybe provide a way to finish it?

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Here's a way to finish the problem after @above's lemma.

Let $P$' be the reflection of $P$ through $O$. Note that $P'$ also satisfies the condition since the perpendiculars through $P$. $P$' to BC are symmetric with respect to the perpendicular bisector of $BC$ if we define $Q'$ similarly , $Q'$ will be the isogonal conjugate of $Q$. Now by lemma above , we can easily see that $Q'$ is in fact one of orthology centers of $\triangle ABC,\triangle DEF$ Click to reveal hidden text

.Now note that $P,P',O$ are collinear and $P',Q',P$ are collinear by Sondat's theorem. Also , $P,Q,Q'$ are collinear , again by Sondat's theorem. So $P,Q,O$ are collinear and we're done.

It is easy to guess that $O$ the center of the circle lay on $PQ$ Let $S$ is the center of the spiral similarity that sends $BC$ to $C_1B_1$. Since $BCC_1B_1$ is cyclic quadrilateral it is known that $S,Q$ and $O$ are colinear. Assume that $R$ is the intersection of $BC$ and $B_1C_1$ then $\angle QSR = 90$. By anglechasing $Q$ is the Isogonal conjugate of $P$ in triangle $A_1B_1C_1$ that means $C_1PB_1$ and $BA_1C$ are similar which means that if you take projections of $P$ on $BC$ and $B_1C_1$ is $X$ and $Y$ then: $\frac{BX}{CX} = \frac{C_1Y}{B_1Y}$ So our spiral sends $X$ to $Y$ which means that $X,Y,S,R$ and $P$ lay on one circle. $\angle PSR = \angle PXR = 90$ So $S,Q,P$ and $O$ are colinear.

We will show that desired point is circumcenter $O$. Define $A_2$ as intersection of $B_1C_1\cap BC$ and similarly define $B_2$ and $C_2$. Define $A_3$ as intersection of $BC_1\cap CB_1$ and similarly define $B_3$ and $C_3$. Since triangles $ABC$ and $A_1B_1C_1$ are perspective, $A_2B_2C_2$ are collinear,call it $\ell$. By Pascal $A_1B_1C_1ABC$ , we see that $A_3$ also lies on $\ell$. Similarly $B_3$ and $C_3$ lies on $\ell$ Notice $ABC$ and $A_1B_1C_1$ are orthologic By Sondat Theorem, orthology centers and perspective center are collinear and this line is perpendicular to perspectrix line(which is $A_2B_2C_2A_3B_3C_3$ in our case). But notice that in quadrilateral $BCB_1C_1$, by Brocard, orthocenter of $A_2A_3Q$ is circumcenter $O$. Both $PQ$ and $OQ$ are perpendicular to line $A_2A_3$, hence $PQO$ are collinear. Q.E.D.