If the point on the curve $$y^2 = 6x$$, nearest to the point $$\left(3, \frac{3}{2}\right)$$ is $$(\alpha, \beta)$$, then $$2(\alpha + \beta)$$ is equal to ___.

We parametrize the parabola $$y^2 = 6x$$ using the standard form with $$4a = 6$$, giving $$a = \frac{3}{2}$$. A general point on the parabola is $$\left(\frac{3t^2}{2},\, 3t\right)$$.

The square of the distance from this point to $$\left(3, \frac{3}{2}\right)$$ is $$D(t) = \left(\frac{3t^2}{2} - 3\right)^2 + \left(3t - \frac{3}{2}\right)^2$$. Differentiating and setting $$D'(t) = 0$$ yields $$9t^3 - 9 = 0$$, so $$t^3 = 1$$, giving the only real solution $$t = 1$$.

At $$t = 1$$, the point on the parabola is $$\alpha = \frac{3(1)^2}{2} = \frac{3}{2}$$ and $$\beta = 3(1) = 3$$. We can verify: $$\beta^2 = 9 = 6 \cdot \frac{3}{2} = 6\alpha$$, confirming the point lies on the parabola.

Therefore $$2(\alpha + \beta) = 2\!\left(\frac{3}{2} + 3\right) = 2 \cdot \frac{9}{2} = \boxed{9}$$.