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

Given: $$\text{Re}\left(\frac{z-\bar{z}+z\bar{z}}{2-3z+5\bar{z}}\right)$$ where $$\text{Re}(z) = 3$$.

Let $$z = 3 + iy$$. Then $$\bar{z} = 3 - iy$$.

Numerator: $$z - \bar{z} + z\bar{z} = 2iy + (9+y^2) = (9+y^2) + 2iy$$.

Denominator: $$2 - 3(3+iy) + 5(3-iy) = 2 - 9 - 3iy + 15 - 5iy = 8 - 8iy$$.

$$\frac{(9+y^2)+2iy}{8-8iy} = \frac{[(9+y^2)+2iy](8+8iy)}{64(1+y^2)}$$

Real part of the numerator product: $$8(9+y^2) + 16i^2y^2 = 8(9+y^2) - 16y^2 = 72 - 8y^2$$.

$$\text{Re} = \frac{72-8y^2}{64(1+y^2)} = \frac{9-y^2}{8(1+y^2)}$$

Let $$f(y) = \frac{9-y^2}{8(1+y^2)}$$. Setting $$f'(y) = 0$$ gives $$y = 0$$ (a maximum). At $$y = 0$$: $$f = 9/8$$. As $$|y| \to \infty$$: $$f \to -1/8$$.

Range of $$f$$: $$(-1/8, 9/8]$$. So $$\alpha = -1/8$$, $$\beta = 9/8$$.

$$24(\beta - \alpha) = 24(9/8 + 1/8) = 24 \times 10/8 = 30$$.

The correct answer is Option 3: $$30$$.