Let the tangent to the parabola $$y^2 = 12x$$ at the point $$(3, \alpha)$$ be perpendicular to the line $$2x + 2y = 3$$. Then the square of distance of the point $$(6, -4)$$ from the normal to the hyperbola $$\alpha^2x^2 - 9y^2 = 9\alpha^2$$ at its point $$(\alpha - 1, \alpha + 2)$$ is equal to _______

The tangent to the parabola $$y^2 = 12x$$ at $$(3, \alpha)$$ is perpendicular to $$2x + 2y = 3$$.

Find $$\alpha$$.

Since $$(3, \alpha)$$ is on $$y^2 = 12x$$: $$\alpha^2 = 36 \implies \alpha = \pm 6$$.

Slope of tangent at $$(3, \alpha)$$: differentiating $$y^2 = 12x$$ gives $$\frac{dy}{dx} = \frac{6}{y} = \frac{6}{\alpha}$$.

Slope of line $$2x + 2y = 3$$ is $$-1$$.

For perpendicularity: $$\frac{6}{\alpha} \times (-1) = -1 \implies \alpha = 6$$.

Set up the hyperbola.

With $$\alpha = 6$$: $$36x^2 - 9y^2 = 9 \times 36$$ simplifies to $$\frac{x^2}{9} - \frac{y^2}{36} = 1$$.

Here $$a^2 = 9, b^2 = 36$$.

Verify the point on the hyperbola.

Point $$(\alpha - 1, \alpha + 2) = (5, 8)$$: $$\frac{25}{9} - \frac{64}{36} = \frac{25}{9} - \frac{16}{9} = 1$$ ✓

Find the normal at $$(5, 8)$$.

Slope of tangent: $$\frac{dy}{dx} = \frac{b^2 x}{a^2 y} = \frac{36 \times 5}{9 \times 8} = \frac{5}{2}$$

Slope of normal: $$-\frac{2}{5}$$

Normal line: $$y - 8 = -\frac{2}{5}(x - 5)$$ which gives $$2x + 5y = 50$$.

Distance from $$(6, -4)$$ to the normal.

$$d = \frac{|2(6) + 5(-4) - 50|}{\sqrt{4 + 25}} = \frac{|12 - 20 - 50|}{\sqrt{29}} = \frac{58}{\sqrt{29}} = \frac{2 \times 29}{\sqrt{29}} = 2\sqrt{29}$$

$$d^2 = 4 \times 29 = 116$$

The answer is $$\boxed{116}$$.