Let the abscissae of the two points $$P$$ and $$Q$$ be the roots of $$2x^2 - rx + p = 0$$ 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 ______.

Let

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

The abscissae of $$P,Q$$ are roots of

$$2x^2-rx+p=0$$

Hence,

$$x_1+x_2=\frac{r}{2},\qquad x_1x_2=\frac{p}{2}$$

Also, the ordinates of $$P,Q$$ are roots of

$$x^2-sx-q=0$$

Hence,

$$y_1+y_2=s,\qquad y_1y_2=-q$$

The equation of the circle having $$PQ$$ as diameter is

$$x^2+y^2-(x_1+x_2)x-(y_1+y_2)y+x_1x_2+y_1y_2=0$$

Substituting the values,

$$x^2+y^2-\frac r2x-sy+\frac p2-q=0$$

The given circle is

$$2(x^2+y^2)-11x-14y-22=0$$

Dividing by $$2,$$

$$x^2+y^2-\frac{11}{2}x-7y-11=0$$

Comparing coefficients,

$$\frac r2=\frac{11}{2},\qquad s=7,\qquad \frac p2-q=-11$$

Hence,

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

Therefore,

$$2r+s-2q+p=2(11)+7+(p-2q)$$

$$=22+7-22$$

$$=7$$

Hence, $$\boxed{7}$$.