As point $P(-2\sqrt{6},\sqrt{3})$ lies on hyperbola.
$\frac{24}{a^2}-\frac{3}{b^2}=1 ...\left(1\right)$

Given, eccentricity $e=\frac{\sqrt{5}}{2}$

$\Rightarrow e=\sqrt{\frac{a^2+b^2}{a^2}}=\frac{\sqrt{5}}{2}$

$\Rightarrow 4a^2+4b^2=5a^2$
$\Rightarrow a^2=4b^2 ...\left(2\right)$

On solving equations $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow \frac{6}{b^2}-\frac{3}{b^2}=1$
$\Rightarrow b^2=3 \& a^2=12$
Hyperbola $\frac{x^2}{12}-\frac{y^2}{3}=1$
$\frac{2x}{12}-\frac{2y·y^{\text{'}}}{3}=0$

$y\text{'}{|}_{(-2\sqrt{6},\sqrt{3})}=-\frac{1}{\sqrt{2}}$

Equation of tangent is $(y-\sqrt{3})=-\frac{1}{\sqrt{2}}(x+2\sqrt{6})$
at $x=0, y=-\sqrt{3}, Q(0,-\sqrt{3})$
Equation of normal is $(y-\sqrt{3})=\sqrt{2}(x+2\sqrt{6})$
at $x=0, y=5\sqrt{3}$
$R(0,5\sqrt{3})$

Apply distance formula,
$QR=6\sqrt{3}$