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

The equation $$\left|\frac{z-2}{z-3}\right| = 2$$ means $$|z - 2| = 2|z - 3|$$.

Let $$z = x + iy$$. Squaring both sides:

$$(x - 2)^2 + y^2 = 4\left[(x - 3)^2 + y^2\right]$$

Expanding: $$x^2 - 4x + 4 + y^2 = 4x^2 - 24x + 36 + 4y^2$$

Simplifying: $$3x^2 + 3y^2 - 20x + 32 = 0$$

Dividing by 3: $$x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0$$

Completing the square for $$x$$:

$$\left(x - \frac{10}{3}\right)^2 + y^2 = \frac{100}{9} - \frac{32}{3} = \frac{100 - 96}{9} = \frac{4}{9}$$

This is a circle with center $$\left(\frac{10}{3},\, 0\right)$$ and radius $$\frac{2}{3}$$.

So $$\alpha = \frac{10}{3}$$, $$\beta = 0$$, and $$\gamma = \frac{2}{3}$$.

$$\alpha + \beta + \gamma = \frac{10}{3} + 0 + \frac{2}{3} = \frac{12}{3} = 4$$

Therefore $$3(\alpha + \beta + \gamma) = 3 \times 4 = 12$$

The answer is Option D: $$12$$.