If the set $\left\{\mathrm{Re}\left(\frac{\mathrm{z}-\overline{\mathrm{z}}+\mathrm{z}\overline{\mathrm{z}}}{2-3\mathrm{z}+5\overline{\mathrm{z}}}\right):\mathrm{z}\in \mathrm{C},\mathrm{Re}\left(\mathrm{z}\right)=3\right\}$ is equal to the interval $(\mathrm{\alpha }, \mathrm{\beta }]$ .Then $24(\mathrm{\beta }-\mathrm{\alpha })$ is equal to

$$\begin{gathered} \begin{array}{l}\mathrm{Re}\left(\frac{z-\overline{z}+z\overline{z}}{2-3z+5\overline{z}}\right)\\ \mathrm{Re}\left(\frac{x+iy-(x-iy)+x^2+y^2}{2-3(x+iy)+5(x-iy)}\right)\\ \mathrm{Re}\left(\frac{x^2+y^2+i\left(2y\right)}{2+2x-8iy}\right) \\ \begin{array}{l}\mathrm{Re}\left(\frac{\left(x^2+y^2+2yi\right)\left(2\right(1+x)+8iy)}{\left(2\right(1+x){)}^2+(8y{)}^2}\right)\\ =\frac{2\left(x^2+y^2\right)(1+x)-16y^2}{4(1+x{)}^2+(8y{)}^2} \\ \text{ Put }x=3 \\ \begin{array}{l}=\frac{8\left(9+y^2\right)-16y^2}{64+64y^2}\\ f\left(y\right)=\frac{1}{8}\frac{\left(9-y^2\right)}{1+y^2}\end{array}\end{array}\end{array} \end{gathered}$$

$$\begin{gathered} \text{ Range of }f\left(y\right)=(-0.125,1.125] \\ \begin{array}{l}\alpha =-0.125\\ \beta =1.125\\ \beta -\alpha =1.25\\ 24(\beta -\alpha )=30\end{array} \end{gathered}$$