The line x = 8 is the directrix of the ellipse $\mathrm{E}:\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ 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 $(3\mathrm{PQ}{)}^2$is equal to _____

$\mathrm{E}\equiv \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 .....\left(1\right)$
Directrix of E : is x = 8 and focus S (2, 0) = (ae, 0)
$\Rightarrow \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=8\dots ..\left(2\right)\mathrm{\&}\mathrm{ae}=2 ....\left(3\right)$
$\begin{array}{r}\text{ (2) }\mathrm{\&}\left(3\right)8{\mathrm{e}}^2=2\\ \Rightarrow \mathrm{e}=1/2,\mathrm{a}=4\end{array}$
$\begin{array}{l}\therefore {\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)={\mathrm{a}}^2-{\mathrm{a}}^2{\mathrm{e}}^2\\ \Rightarrow {\mathrm{b}}^2=16-4\\ \Rightarrow {\mathrm{b}}^2=12\\ \therefore \mathrm{E}\equiv \frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{12}=1\end{array}$

Equation of tangent of $\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{ to }\mathrm{E}=0\text{, is }{\mathrm{S}}_1=0$
$\text{ i.e }\frac{{\mathrm{xx}}_1}{16}+\frac{{\mathrm{yy}}_1}{12}=1$
passing through the point $\begin{array}{r}(0,4\sqrt{3}),\text{ so }\frac{{\mathrm{y}}_1\left(4\sqrt{3}\right)}{12}=1\\ \Rightarrow {\mathrm{y}}_1=\sqrt{3}\end{array}$
So point on $\frac{{\mathrm{x}}_1^2}{16}+\frac{{\mathrm{y}}_1^2}{12}=1$
So equation of tangent at $\mathrm{P}(2\sqrt{3},\sqrt{3})\text{ of }\mathrm{E}=0,{\mathrm{S}}_1=0$
i.e.,  $3\mathrm{x}+2\mathrm{y}=8\sqrt{3}$ interest x-axis at $\mathrm{Q}\left(\frac{8}{\sqrt{3}},0\right) \therefore {\mathrm{PQ}}^2=\frac{13}{3}$
$(3\mathrm{PQ}{)}^2=9(\mathrm{PQ}{)}^2=9\left(\frac{13}{3}\right)=39$