If the set $\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in \mathbb{C}, \operatorname{Re}(z)=3\right\}$ is equal to

the interval $(\alpha, \beta]$, then $24(\beta-\alpha)$ is equal to :

Let $z_1=\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right)$<br/><br/> Let $\mathrm{z}=3+\mathrm{iy}$<br/><br/> $\bar{z}=3-i y$<br/><br/> $$ \begin{aligned} & z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i y)+5(3-i y)} \\\\ & =\frac{9+y^2+i(2 y)}{8-8 i y} \\\\ & =\frac{\left(9+y^2\right)+i(2 y)}{8(1-i y)} \\\\ & \operatorname{Re}\left(z_1\right)=\frac{\left(9+y^2\right)-2 y^2}{8\left(1+y^2\right)} \end{aligned} $$<br/><br/> $$ \begin{aligned} & =\frac{9-y^2}{8\left(1+y^2\right)} \\\\ & =\frac{1}{8}\left[\frac{10-\left(1+y^2\right)}{\left(1+y^2\right)}\right] \\\\ & =\frac{1}{8}\left[\frac{10}{1+y^2}-1\right] \\\\ & 1+y^2 \in[1, \infty] \\\\ & \frac{1}{1+y^2} \in(0,1] \\\\ & \frac{10}{1+y^2} \in(0,10] \\\\ & \frac{10}{1+y^2}-1 \in(-1,9] \\\\ & \operatorname{Re}\left(\mathrm{z}_1\right) \in\left(\frac{-1}{8}, \frac{9}{8}\right] \\\\ & \alpha=\frac{-1}{8}, \beta=\frac{9}{8} \\\\ & 24(\beta-\alpha)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30 \end{aligned} $$