Solution
Let $\mathrm{z}_{1}=\left(\frac{\mathrm{z}-\overline{\mathrm{z}}+\mathrm{z} \overline{\mathrm{z}}}{2-3 \mathrm{z}+5 \overline{\mathrm{z}}}\right)$ Let $\mathrm{z}=3+\mathrm{i} y, \overline{\mathrm{z}}=3-\mathrm{iy} \Rightarrow \mathrm{z}_{1}=\frac{2 \mathrm{iy}+\left(9+\mathrm{y}^{2}\right)}{2-3(3+\mathrm{iy})+5(3-\mathrm{iy})}=\frac{9+\mathrm{y}^{2}+\mathrm{i}(2 \mathrm{y})}{8-8 \mathrm{iy}}=\frac{\left(9+\mathrm{y}^{2}\right)+\mathrm{i}(2 \mathrm{y})}{8(\mathrm{i}-\mathrm{iy})}$
\[
\begin{aligned}
& \operatorname{Re}\left(z_{1}\right)=\frac{\left(9+y^{2}\right)-2 y^{2}}{8\left(i+y^{2}\right)}=\frac{9-y^{2}}{8\left(i+y^{2}\right)}=\frac{1}{8}\left[\frac{10}{i+y^{2}}-1\right] \because 1+y^{2} \in[1, \infty] \Rightarrow \frac{1}{i+y^{2}} \in(0,1] \Rightarrow \frac{10}{i+y^{2}}-1 \in(-1,9] \\
& \operatorname{Re}\left(z_{1}\right) \in\left(-\frac{1}{8}, \frac{9}{8}\right] \Rightarrow \alpha=-\frac{1}{8}, \beta=\frac{9}{8} \Rightarrow 24(\alpha-\beta)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30
\end{aligned}
\]
.
