Let the abscissae of the two points P and Q be the roots of $2{x^2} - rx + p = 0$ and the ordinates of P and Q be the roots of ${x^2} - sx - q = 0$. If the equation of the circle described on PQ as diameter is $2({x^2} + {y^2}) - 11x - 14y - 22 = 0$, then $2r + s - 2q + p$ is equal to __________.

<p>Let $P({x_1},{y_1})$ & $Q({x_2},{y_2})$</p> <p>$\therefore$ Roots of $2{x^2} - rx + p = 0$ are ${x_1},\,{x_2}$</p> <p>and roots of ${x^2} - sx - q = 0$ are ${y_1},\,{y_2}$.</p> <p>$\therefore$ Equation of circle $\equiv (x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0$</p> <p>$$ \Rightarrow {x^2} - ({x_1} + {x_2})x + {x_1}{x_2} + {y^2} - ({y_1} + {y_2})y + {y_1}{y_2} = 0$$</p> <p>$\Rightarrow {x^2} - {r \over 2}x + {p \over 2} + {y^2} + sy - q = 0$</p> <p>$\Rightarrow 2{x^2} + 2{y^2} - rx + 2sy + p - 2q = 0$</p> <p>Compare with $2{x^2} + 2{y^2} - 11x - 14y - 22 = 0$</p> <p>We get $r = 11,\,s = 7,\,p - 2q = - 22$</p> <p>$\Rightarrow 2r + s + p - 2q = 22 + 7 - 22 = 7$</p>