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 ______.

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

First, to determine $$a$$ and $$b$$, recall that for an ellipse the focus is at $$(ae, 0)$$ and the corresponding directrix is at $$x = \frac{a}{e}$$, so we have $$ ae = 2 \quad \text{and} \quad \frac{a}{e} = 8 $$. Multiplying these gives $$a^2 = 16$$, hence $$a = 4$$.

Since $$ae = 2$$, it follows that $$e = \frac{1}{2}$$. Then using $$b^2 = a^2(1 - e^2) = 16\left(1 - \frac{1}{4}\right) = 12$$, the equation of the ellipse becomes $$\frac{x^2}{16} + \frac{y^2}{12} = 1$$.

Next, to find the point $$P$$ in the first quadrant, note that the equation of the tangent to the ellipse at $$P(x_0, y_0)$$ is $$ \frac{x \cdot x_0}{16} + \frac{y \cdot y_0}{12} = 1 $$ and it passes through $$(0, 4\sqrt{3})$$. Substituting yields $$ \frac{0}{16} + \frac{4\sqrt{3} \cdot y_0}{12} = 1 $$, so $$ y_0 = \frac{12}{4\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} $$.

Because $$P$$ lies on the ellipse, $$\frac{x_0^2}{16} + \frac{3}{12} = 1 \Rightarrow \frac{x_0^2}{16} = \frac{3}{4} \Rightarrow x_0^2 = 12 \Rightarrow x_0 = 2\sqrt{3}$$, giving $$P = (2\sqrt{3}, \sqrt{3})$$.

To find the x-intercept $$Q$$ of the tangent, set $$y = 0$$ in the tangent equation: $$ \frac{x \cdot 2\sqrt{3}}{16} = 1 \Rightarrow x = \frac{16}{2\sqrt{3}} = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3} $$. Thus $$Q = \left(\frac{8\sqrt{3}}{3}, 0\right)$$.

Finally, to compute $$(3PQ)^2$$, first evaluate $$ PQ^2 = \left(2\sqrt{3} - \frac{8\sqrt{3}}{3}\right)^2 + (\sqrt{3} - 0)^2 $$ which gives $$ = \left(\frac{6\sqrt{3} - 8\sqrt{3}}{3}\right)^2 + 3 = \left(\frac{-2\sqrt{3}}{3}\right)^2 + 3 $$. Continuing, $$ = \frac{12}{9} + 3 = \frac{4}{3} + 3 = \frac{13}{3} $$. Therefore, $$(3PQ)^2 = 9 \cdot PQ^2 = 9 \times \frac{13}{3} = 39$$.