Let $P=(x_1,y_1),\, Q=(x_2,y_2).$

We are given:

• the abscissae $x_1,x_2$ are roots of $2x^2-rx+p=0,$
• the ordinates $y_1,y_2$ are roots of $x^2-sx-q=0.$

The circle with diameter $PQ$ is $2(x^2+y^2)-11x-14y-22=0.$

1. Write the circle in standard form

Divide by $2$: $x^2+y^2-\frac{11}{2}x-7y-11=0.$

Compare with the general equation of a circle having diameter endpoints $(x_1,y_1)$ and $(x_2,y_2)$: $x^2+y^2-(x_1+x_2)x-(y_1+y_2)y+x_1x_2+y_1y_2=0.$

So we get $x_1+x_2=\frac{11}{2}, \qquad y_1+y_2=7, \qquad x_1x_2+y_1y_2=-11.$

2. Use Vieta's formulas for the abscissae

For $2x^2-rx+p=0,$ the sum and product of roots are $x_1+x_2=\frac{r}{2}, \qquad x_1x_2=\frac{p}{2}.$

Since $x_1+x_2=\frac{11}{2}$, $\frac{r}{2}=\frac{11}{2} \implies r=11.$

3. Use Vieta's formulas for the ordinates

For $x^2-sx-q=0,$ the sum and product of roots are $y_1+y_2=s, \qquad y_1y_2=-q.$

Since $y_1+y_2=7$, $s=7.$

4. Use the constant term of the circle

We have $x_1x_2+y_1y_2=-11.$

Now $x_1x_2=\frac{p}{2}, \qquad y_1y_2=-q.$

Hence $\frac{p}{2}-q=-11.$ Multiply by $2$: $p-2q=-22.$

5. Find the required expression

We need $2r+s-2q+p.$

Substitute $r=11$, $s=7$, and $p-2q=-22$: $2r+s-2q+p=2(11)+7+(p-2q)=22+7-22=7.$

Final Answer

$\boxed{7}$

The derived answer matches the stored correct answer.