Let $H$ denote the orthocenter of $\triangle ABC$. We claim that $\ell$ always passes through $H$.

Lemma. Line $PQ$ bisects segment $XH$.

Proof. Let $X_A, X_B$ be the reflections of $X$ across $BC$ and $AC$ respectively, and let $H_A$ be the reflection of $H$ across $BC$. It is easy to see that since $\angle BH_A C = \angle BHC = 180^\circ - \angle BAC$, $H_A$ is on the circumcircle of $\triangle ABC$. It suffices to show that $H$ is on $X_A X_B$. Since $C$ is the circumcenter of $XX_A X_B$, we have $\angle XX_A X_B = \frac{1}{2} \angle XCX_B = \angle ACX$. On the other hand, $HH_A X_A X$ is an isosceles trapezoid, so $\angle HX_A X = \angle HH_A X = \angle AH_A X = \angle ACX = \angle XX_A X_B$ so it follows that $X_B, H, X_A$ are collinear.

We know that lines $HBR$ and $XP$ are both perpendicular to $AC$, so it follows that $HR \parallel XP$. But by the lemma, line $PQR$ bisects $HX$ so it follows that $PXRH$ is a parallelogram. Thus, $PH \parallel XR$ and thus $H$ is on $\ell$, as desired.