Given that

$z \neq \overline{ z }$

Let $\alpha=\frac{2+3 z+4 z^2}{2-3 z+4 z^2}=\frac{\left(2-3 z+4 z^2\right)+6 z}{2-3 z+4 z^2}$

$\therefore \alpha=1+\frac{6 z}{2-3 z+4 z^2}$

If $\alpha$ is a real number, then

$\alpha=\bar{\alpha}$

$\Rightarrow \frac{z}{2-3 z+4 z^2}=\frac{\bar{z}}{2-3 \bar{z}+4 \bar{z}^2}$

$\therefore 2(z-\bar{z})=4 z \bar{z}(z-\bar{z})$

$\Rightarrow(z-\bar{z})(2-4 z \bar{z})=0$

As $z \neq \bar{Z}$ (Given)

$\Rightarrow z \bar{z}=\frac{2}{4}=\frac{1}{2}$

$\Rightarrow|z|^2=0.50$