Let $$P({x_1},{y_1})$$ & $$Q({x_2},{y_2})$$

$$\therefore$$ Roots of $$2{x^2} - rx + p = 0$$ are $${x_1},\,{x_2}$$

and roots of $${x^2} - sx - q = 0$$ are $${y_1},\,{y_2}$$.

$$\therefore$$ Equation of circle $$ \equiv (x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0$$

$$ \Rightarrow {x^2} - ({x_1} + {x_2})x + {x_1}{x_2} + {y^2} - ({y_1} + {y_2})y + {y_1}{y_2} = 0$$

$$ \Rightarrow {x^2} - {r \over 2}x + {p \over 2} + {y^2} + sy - q = 0$$

$$ \Rightarrow 2{x^2} + 2{y^2} - rx + 2sy + p - 2q = 0$$

Compare with $$2{x^2} + 2{y^2} - 11x - 14y - 22 = 0$$

We get $$r = 11,\,s = 7,\,p - 2q = - 22$$

$$ \Rightarrow 2r + s + p - 2q = 22 + 7 - 22 = 7$$