Let the roots of 2x2−rx+p=0 are x1,x2 and roots of y2−sy−q=0 are y1,y2
So, x1+x2=2r,x1x2=2p,y1+y2=s,y1y2=−q
Equation of the circle with PQ as diameter will be
(x−x1)(x−x2)+(y−y1)(y−y2)=0
i.e. 2(x2+y2)−rx−2sy+p−2q=0
On comparing with the given equation r=11,s=7
p−2q=−22
∴2r+s−2q+p=22+7−22=7