$\left(\frac{z^2+81 z-15}{z^2-3 i z-2}\right) \in R$

$\Rightarrow 1+\frac{(11 i z-13)}{\left(z^2-3 i z-2\right)} \in R$

Put $z=\alpha-\frac{13}{11} i$

$\Rightarrow\left(z^2-3 i z-2\right)$ is imaginary

Put $z=x+i y$

$\Rightarrow\left(x^2-y^2+2 x y i-3 i x+3 y-2\right) \in \text { Imaginary }$

$\Rightarrow \operatorname{Re}\left(x^2-y^2+3 y-2+(2 x y-3 x) i\right)=0$

$\Rightarrow x^2-y^2+3 y-2=0$

$x^2=y^2-3 y+2$

$x^2=y^2-3 y+2$

$x^2=(y-1)(y-2) \therefore z=\alpha-\frac{13}{11} i$

Put $x=\alpha, y=\frac{-13}{11}$

$\alpha^2=\left(\frac{-13}{11}-1\right)\left(\frac{-13}{11}-2\right)$

$\alpha^2=\frac{(24 \times 35)}{121}$

$242 \alpha^2=48 \times 35=1680$