1. Parametrize the parabola

The parabola is $y^2 = 6x.$ This matches the standard form $y^2 = 4ax$ with $4a = 6 \implies a = \frac{3}{2}.$

A general point on the parabola can be written as $\left(at^2,\, 2at\right).$ So here, $x = \frac{3}{2}t^2, \qquad y = 3t.$

Thus a variable point on the parabola is $P\left(\frac{3}{2}t^2,\, 3t\right).$

The given point is $A\left(3, \frac{3}{2}\right).$

2. Write the squared distance

To find the nearest point, minimize the square of the distance: $D^2 = \left(\frac{3}{2}t^2 - 3\right)^2 + \left(3t - \frac{3}{2}\right)^2.$

Factor out $\left(\frac{3}{2}\right)^2$ if desired: $D^2 = \left(\frac{3}{2}\right)^2\left[(t^2-2)^2 + (2t-1)^2\right].$

So we minimize $f(t) = (t^2-2)^2 + (2t-1)^2.$

3. Differentiate and set equal to zero

Expand first: $f(t) = t^4 - 4t^2 + 4 + 4t^2 - 4t + 1 = t^4 - 4t + 5.$

Now differentiate: $f'(t) = 4t^3 - 4.$

Set this equal to zero: $4t^3 - 4 = 0$ $t^3 = 1$ $t = 1.$

Also, $f''(t) = 12t^2 > 0$ for $t=1$, so this gives a minimum.

4. Find the nearest point

Substitute $t=1$ into the parametric coordinates: $\alpha = \frac{3}{2}(1)^2 = \frac{3}{2},$ $\beta = 3(1) = 3.$

So the nearest point is $\left(\frac{3}{2}, 3\right).$

5. Compute the required value

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

Final Answer

$\boxed{9}$

6. Comparison with stored answer

Stored correct answer = $9$.

Our derived answer is also $9$, so they agree.