There are certain aspects to this question requiring clarification.
However, it's possible to produce a solution using three sets of coordinates for the vertices of PQR:
P(Px,Py), Q(Qx,Qy), R(Rx,Ry).
From these the midpoint of PQ is F(½(Px+Qx),½(Py+Qy).
(a)
Using Pythagoras' Theorem:
RP2=(Rx-Px)2+(Ry-Py)2; FR2=(½(Px+Qx)-Rx)2+(½(Py+Qy)-Ry)2.
RP=√[(Rx-Px)2+(Ry-Py)2], FR=√[(½(Px+Qx)-Rx)2+(½(Py+Qy)-Ry)2].
(b)
PQ=√[(Px-Qx)2+(Py-Qy)2]; QR=√[(Qx-Rx)2+(Qy-Ry)2].
The proposition 2PQ+3QR+RP=2FR must apply generally to all triangles because no specific numerical lengths have been provided. So if we take a right triangle PQR such that PQ=10, QR=6 and PR=8, PF=FQ=5. 2PQ+3QR+RP=20+18+8=46, but by simple geometry, FR=5, so 2FR=10 and the proposition is clearly false. Since there is no given relation between w, x, y, z as independent variables, and no other constraints have been mentioned, some values assigned to the variables could produce the right triangle example, whatever these variables indicate. Nevertheless, we can derive relationships as follows:
Sine Rule:
sinQP̂R/QR=sinPQ̂R/RP=sinQR̂P/PQ (triangle PQR);
sinQP̂R/FR=sinPF̂R/RP=sinFR̂P/PF (triangle PFR) (QP̂R and FP̂R are the same angle);
sinQF̂R/QR=sinPQ̂R/FR=sinQR̂F/FQ (triangle FQR) (PQ̂R and FQ̂R are the same angle).
And PF=FQ, FR̂P+QR̂F=QR̂P. Let a=PF=FQ; let b=QR̂P and c=FR̂P. (Lengths of sides and measures of angles.)
sinQR̂F=sin(b-c)=sin(b)cos(c)-cos(b)sin(c).
FR is common to triangles FQR and PFR, and PF=FQ=a, PQ=2a. The LHS of the proposed equation contains the green highlighted lengths. The other highlighting may help to understand the following logic.
More may follow in due course after clarification...