Let the abscissae of the two points P and Q be the roots of 2 x ^ 2...

Let the roots of $2{x}^{2} - rx + p = 0$ are ${x}_{1}, {x}_{2} and roots of${y}^{2} - sy - q = 0$are${y}_{1}, {y}_{2}$</p><p>So,${x}_{1} + {x}_{2} = \dfrac{r}{2}, {x}_{1}{x}_{2} = \dfrac{p}{2}, {y}_{1} + {y}_{2} = s, {y}_{1}{y}_{2} = - q$</p><p>Equation of the circle with$PQ$as diameter will be </p><p>\left(x - {x}_{1}\right)\left(x - {x}_{2}\right) + \left(y - {y}_{1}\right)\left(y - {y}_{2}\right) = 0$

i.e. $2\left({x}^{2} + {y}^{2}\right) - rx - 2sy + p - 2q = 0$

On comparing with the given equation $r = 11, s = 7$

$p - 2q = - 22$

$∴ 2r + s - 2q + p = 22 + 7 - 22 = 7$