Let the abscissae of the two points $$P$$ and $$Q$$ on a circle be the roots of $$x^2 - 4x - 6 = 0$$ and the ordinates of $$P$$ and $$Q$$ be the roots of $$y^2 + 2y - 7 = 0$$. If $$PQ$$ is a diameter of the circle $$x^2 + y^2 + 2ax + 2by + c = 0$$, then the value of $$a + b - c$$ is

We are given that the abscissae of points $$P$$ and $$Q$$ are roots of $$x^2 - 4x - 6 = 0$$ and the ordinates are roots of $$y^2 + 2y - 7 = 0$$. $$PQ$$ is a diameter of the circle $$x^2 + y^2 + 2ax + 2by + c = 0$$.

Since $$PQ$$ is a diameter, the center of the circle is the midpoint of $$PQ$$.

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

From $$x^2 - 4x - 6 = 0$$: $$x_1 + x_2 = 4$$ (sum of roots)

From $$y^2 + 2y - 7 = 0$$: $$y_1 + y_2 = -2$$ (sum of roots)

Center = $$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) = (2, -1)$$

Comparing with the general form $$x^2 + y^2 + 2ax + 2by + c = 0$$, the center is $$(-a, -b)$$.

$$-a = 2 \Rightarrow a = -2$$

$$-b = -1 \Rightarrow b = 1$$

$$|PQ|^2 = (x_1-x_2)^2 + (y_1-y_2)^2$$

$$(x_1-x_2)^2 = (x_1+x_2)^2 - 4x_1x_2 = 16 - 4(-6) = 16 + 24 = 40$$

$$(y_1-y_2)^2 = (y_1+y_2)^2 - 4y_1y_2 = 4 - 4(-7) = 4 + 28 = 32$$

$$|PQ|^2 = 40 + 32 = 72$$

Radius$$^2 = \frac{|PQ|^2}{4} = \frac{72}{4} = 18$$

For the circle $$x^2 + y^2 + 2ax + 2by + c = 0$$:

$$r^2 = a^2 + b^2 - c$$

$$18 = (-2)^2 + 1^2 - c = 4 + 1 - c = 5 - c$$

$$c = 5 - 18 = -13$$

$$a + b - c = -2 + 1 - (-13) = -2 + 1 + 13 = 12$$

Therefore, the correct answer is Option A: $$12$$.