A line is a common tangent to the circle $$(x - 3)^2 + y^2 = 9$$ and the parabola $$y^2 = 4x$$. If the two points of contact $$(a, b)$$ and $$(c, d)$$ are distinct and lie in the first quadrant, then $$2(a + c)$$ is equal to ______.

The tangent to the parabola $$y^2 = 4x$$ at the point $$(t^2, 2t)$$ is given by $$ty = x + t^2$$, which can be written as $$x - ty + t^2 = 0$$.

For this line to be tangent to the circle $$(x-3)^2 + y^2 = 9$$ (center $$(3, 0)$$, radius $$3$$), the perpendicular distance from $$(3, 0)$$ to the line must equal 3: $$\frac{|3 - 0 + t^2|}{\sqrt{1 + t^2}} = 3$$.

Since $$t^2 + 3 > 0$$, we square both sides: $$(3 + t^2)^2 = 9(1 + t^2)$$. Expanding: $$9 + 6t^2 + t^4 = 9 + 9t^2$$, which gives $$t^4 - 3t^2 = 0$$, so $$t^2(t^2 - 3) = 0$$. Thus $$t^2 = 3$$, and for the first quadrant point on the parabola, $$t = \sqrt{3}$$.

The tangent line is $$\sqrt{3}\,y = x + 3$$. The point of contact on the parabola is $$(t^2, 2t) = (3, 2\sqrt{3})$$, so $$c = 3$$ and $$d = 2\sqrt{3}$$.

For the point of contact on the circle, we find the foot of the perpendicular from the center $$(3, 0)$$ to the line $$x - \sqrt{3}\,y + 3 = 0$$. Using the formula: $$(a, b) = \left(3 - \frac{3 + 0 + 3}{1 + 3} \cdot 1,\; 0 - \frac{3 + 0 + 3}{1 + 3} \cdot (-\sqrt{3})\right) = \left(3 - \frac{3}{2},\; \frac{3\sqrt{3}}{2}\right) = \left(\frac{3}{2},\; \frac{3\sqrt{3}}{2}\right)$$.

So $$a = \dfrac{3}{2}$$. Therefore, $$2(a + c) = 2\left(\dfrac{3}{2} + 3\right) = 2 \times \dfrac{9}{2} = 9$$.