We have: $ \left\| \begin{array}{cc} BPQC \rightarrow \angle{PBQ} = \angle{PCQ} \\ AQSB \rightarrow \angle{ABQ} = \angle{ASQ} \\ APRC \rightarrow \angle{PCA} = \angle{PRA} \end{array} \right\| \Rightarrow \angle{ABQ} = \angle{ASQ} = \angle{PCA} = \angle{PRA}$ (*1) Also $ \left\| \begin{array}{cc} AQSB \rightarrow \angle{A} = 180^\circ - \angle{BSQ} = \angle{QSC} \\ APRC \rightarrow \angle{A} = 180^\circ - \angle{PRC} = \angle{PRB} \end{array} \right\| \Rightarrow \angle{QSC} = \angle{PRB}$ (*2) From (*1) and (*2) obtain that triangles $ ARS$ and $ RLS$ are isosceles, so quadrilateral $ ARLS$ is kite-shaped and $ AL\perp BC$. From here their intersection is always constant (foot of the altitude from $ A$).

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I´ve found another solution: Let D, E and F the foot of the altitude from A, B and C respectively. Now, it´s easy to see that DE//QS; because both are antiparalells to BC. Analogly, we have DF//PR and EF//PQ. Then DEF and LPQ are homothetic from A, then, A, D and L are collinears. Now we can conclude like the prove above.

here is the sketch for another proof Let $ P', Q'$ be another pair of points like $ P, Q$. In the same way define $ L'$. Then: the sides of the triangle $ PQL$ are parallel to the sides of the triangle $ P'Q'L'$ Let $ T = AL\cap BC$, $ U\in AC$ such that $ UT||LQ$ and $ V\in AB$ such that $ VT||AC$. Then: the sides of triangle $ UVT$ are parallel to the sides of triangle $ PQL$ Prove that there exist only one triangle with its vertex on the sides of $ ABC$ with sides parallel to a given reference triangle. This proves that triangle $ UVT$ is "constant", in particular $ T$ is the always the same point.

let $ \angle CAL = \alpha$. note that $ \angle AQL = \pi - B$ and $ \angle APL = \pi - C$, therefore $ \dfrac{\sin\alpha}{\sin (A - \alpha)} = \dfrac{QL\sin B}{PL\sin C}$ note that $ \angle PQL = \pi - 2B$ and $ \angle QPL = \pi - 2C$, so, $ \dfrac{\sin\alpha}{\sin (A - \alpha)} = \dfrac{\sin 2C\sin B}{\sin 2B\sin C} = \dfrac{\cos C}{\cos B}$... given that $ \dfrac{\sin x}{\sin (A - x)}$ is an increasing function, it follows that $ \alpha$ is constant, which is equivalent to the problem statement... in fact, it follows from $ \dfrac{\sin\alpha}{\sin (A - \alpha)} = \dfrac{\cos C}{\cos B}$ that $ \alpha = \frac {\pi}{2} - C$, which implies that $ AL\perp BC$.

This problem is mine, my original proof is the following step1: Triangle LSR is isosceles step2: A is the L-excenter of triangle PQL, which implies AL bisects angle SLR step3: conclude that AL is perpendicular to BC.