The line $$x = 8$$ is the directrix of the ellipse $$E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with the corresponding focus $$(2, 0)$$. If the tangent to $$E$$ at the point $$P$$ in the first quadrant passes through the point $$(0, 4\sqrt{3})$$ and intersects the $$x$$-axis at $$Q$$, then $$(3PQ)^2$$ is equal to _____.

Given the ellipse $$ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ with directrix $$ x = 8 $$ and corresponding focus $$ (2, 0) $$.

The center of the ellipse is at $$ (0,0) $$. The distance from the center to the focus is $$ ae = 2 $$. The distance from the center to the directrix is $$ \frac{a}{e} = 8 $$.

Solving these equations:

Multiply $$ ae $$ and $$ \frac{a}{e} $$: $$ (ae) \times \left( \frac{a}{e} \right) = 2 \times 8 $$ ⇒ $$ a^2 = 16 $$ ⇒ $$ a = 4 $$ (since $$ a > 0 $$).

Substitute $$ a = 4 $$ into $$ ae = 2 $$: $$ 4e = 2 $$ ⇒ $$ e = \frac{1}{2} $$.

Now, $$ b^2 = a^2 (1 - e^2) = 16 \left(1 - \left(\frac{1}{2}\right)^2\right) = 16 \left(1 - \frac{1}{4}\right) = 16 \times \frac{3}{4} = 12 $$.

Thus, the ellipse equation is $$ \frac{x^2}{16} + \frac{y^2}{12} = 1 $$.

The tangent to the ellipse at point $$ P(x_1, y_1) $$ in the first quadrant is given by $$ \frac{x x_1}{16} + \frac{y y_1}{12} = 1 $$. This tangent passes through $$ (0, 4\sqrt{3}) $$.

Substitute $$ x = 0 $$, $$ y = 4\sqrt{3} $$: $$ \frac{0 \cdot x_1}{16} + \frac{(4\sqrt{3}) y_1}{12} = 1 $$ ⇒ $$ \frac{4\sqrt{3} y_1}{12} = 1 $$ ⇒ $$ \frac{\sqrt{3} y_1}{3} = 1 $$ ⇒ $$ \sqrt{3} y_1 = 3 $$ ⇒ $$ y_1 = \sqrt{3} $$.

Since $$ P $$ lies on the ellipse: $$ \frac{x_1^2}{16} + \frac{(\sqrt{3})^2}{12} = 1 $$ ⇒ $$ \frac{x_1^2}{16} + \frac{3}{12} = 1 $$ ⇒ $$ \frac{x_1^2}{16} + \frac{1}{4} = 1 $$ ⇒ $$ \frac{x_1^2}{16} = \frac{3}{4} $$ ⇒ $$ x_1^2 = 12 $$ ⇒ $$ x_1 = 2\sqrt{3} $$ (since $$ P $$ is in the first quadrant).

Thus, $$ P(2\sqrt{3}, \sqrt{3}) $$.

The tangent at $$ P $$ is $$ \frac{x (2\sqrt{3})}{16} + \frac{y (\sqrt{3})}{12} = 1 $$, which simplifies to:

$$ \frac{2\sqrt{3} x}{16} + \frac{\sqrt{3} y}{12} = 1 $$ ⇒ $$ \sqrt{3} \left( \frac{x}{8} + \frac{y}{12} \right) = 1 $$ ⇒ $$ \frac{x}{8} + \frac{y}{12} = \frac{1}{\sqrt{3}} $$.

Multiply through by 24 to clear denominators: $$ 24 \times \frac{x}{8} + 24 \times \frac{y}{12} = 24 \times \frac{1}{\sqrt{3}} $$ ⇒ $$ 3x + 2y = \frac{24}{\sqrt{3}} = \frac{24 \sqrt{3}}{3} = 8\sqrt{3} $$.

So the tangent equation is $$ 3x + 2y = 8\sqrt{3} $$.

This tangent intersects the x-axis at $$ Q $$, where $$ y = 0 $$:

$$ 3x + 2(0) = 8\sqrt{3} $$ ⇒ $$ 3x = 8\sqrt{3} $$ ⇒ $$ x = \frac{8\sqrt{3}}{3} $$.

Thus, $$ Q\left( \frac{8\sqrt{3}}{3}, 0 \right) $$.

Now, find the distance $$ PQ $$ between $$ P(2\sqrt{3}, \sqrt{3}) $$ and $$ Q\left( \frac{8\sqrt{3}}{3}, 0 \right) $$:

$$ \Delta x = \frac{8\sqrt{3}}{3} - 2\sqrt{3} = \frac{8\sqrt{3}}{3} - \frac{6\sqrt{3}}{3} = \frac{2\sqrt{3}}{3} $$,

$$ \Delta y = 0 - \sqrt{3} = -\sqrt{3} $$.

$$ PQ = \sqrt{ (\Delta x)^2 + (\Delta y)^2 } = \sqrt{ \left( \frac{2\sqrt{3}}{3} \right)^2 + (-\sqrt{3})^2 } = \sqrt{ \frac{4 \times 3}{9} + 3 } = \sqrt{ \frac{12}{9} + \frac{27}{9} } = \sqrt{ \frac{39}{9} } = \sqrt{ \frac{13}{3} } = \frac{\sqrt{39}}{3} $$.

Now, compute $$ 3PQ $$:

$$ 3PQ = 3 \times \frac{\sqrt{39}}{3} = \sqrt{39} $$.

Thus, $$ (3PQ)^2 = (\sqrt{39})^2 = 39 $$.

Therefore, $$ (3PQ)^2 = 39 $$.