If the center and radius of the circle $$\left|?\frac{z-2}{z-3} \right| = 2$$ are respectively $$(?lpha, ?eta)$$ and $$\gamma$$, then $$3(?lpha + ?eta + \gamma)$$ is equal to

We need to find the center $$(\alpha, \beta)$$ and radius $$\gamma$$ of the circle $$\left|\frac{z-2}{z-3}\right| = 2$$, then compute $$3(\alpha + \beta + \gamma)$$.

Let $$z = x + iy$$. The equation $$\left|\frac{z-2}{z-3}\right| = 2$$ means $$|z - 2| = 2|z - 3|$$. Squaring both sides gives $$(x-2)^2 + y^2 = 4\bigl[(x-3)^2 + y^2\bigr].$$

Expanding and simplifying yields $$x^2 - 4x + 4 + y^2 = 4x^2 - 24x + 36 + 4y^2,$$ which rearranges to $$3x^2 + 3y^2 - 20x + 32 = 0.$$ Dividing by 3 gives $$x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0.$$

Completing the square, we write $$\left(x - \frac{10}{3}\right)^2 - \frac{100}{9} + y^2 + \frac{32}{3} = 0,$$ so $$\left(x - \frac{10}{3}\right)^2 + y^2 = \frac{100}{9} - \frac{96}{9} = \frac{4}{9}.$$

Thus the center is $$(\alpha, \beta) = \left(\frac{10}{3}, 0\right)$$ and the radius is $$\gamma = \sqrt{\frac{4}{9}} = \frac{2}{3}$$.

Finally, $$3(\alpha + \beta + \gamma) = 3\!\left(\frac{10}{3} + 0 + \frac{2}{3}\right) = 3 \times \frac{12}{3} = 3 \times 4 = 12.$$ The answer is Option D: $$12$$.