
By observation we know the point A(α,0) and β=y-coordinate of R=2+12×4+1×0=38⋯(1)
Now P′ is image of P in y=0 which will be P′(2,−3)
∴ Equation of P′Q is (y+3)=5−24+3(x−2)
i.e. 3y+9=7x−14
Now P′Q meets x−axis at A
so A≡(723,0) by solving with y=0
∴α=723⋯(2)
By (1),(2), we get
7α+3β=23+8=31