Given conic is $\frac{x^2}{12}+\frac{y^2}{16}=1$

$e= \sqrt{1-\frac{12}{16}}=\frac{1}{2}$

Foci $\left(0,2\right)$ and $\left(0,-2\right)$.

So, transverse axis of hyperbola$=2b=4$.

$\Rightarrow$$b=2$

Given $e=\frac{3}{2}$

And $a^2=b^2\left(e^2-1\right)$

$\Rightarrow$$a^2=4 \left(\frac{9}{4}-1\right)$

$\Rightarrow$$a^2=5$

$\therefore$The equation of hyperbola is $\frac{x^2}{5}-\frac{y^2}{4}=-1$.

Now, Check options.

Clearly, the point $\left(5,2\sqrt{3}\right)$ does not satisfy above equation.