<Tri-2023-2072>, hint

Let I. I'be the incenter of triangles ABC. ADE, which are similar. and T be the intesection of DE and AK ( A,I,K are collinear)

 Apply Menelaus theorem

Hence m to triangle ALT with traverse PI'Q.

(AQ/QL)x(LP/PT)x(TI'/I'A)=1. 

Since TI'/I'A=KI/IA ( Triangles ADE and ABC are simialr).  Let A' be antipode of A

LP/PT=AA'/(AA'-AI). AA'=2R sin(B+A/2), AI=(b+c-a/2)/cosA/2

AA'=2R cos(B-C/2), AI=2R (cos(B-C)/2-cos<B+C)2. AA'/(AA'-AI)=cos(B+C/2)/cos(B-C)/2

LP/PT=(b+c)/a=(cos(B-C)/2)/(cos(B+C)/2). Hence Q I' P are collinear.