Given:

$\left|\frac{z-2}{z-3}\right|=2$

Let $z=x+iy$, then we have

$\left|\frac{x-2+iy}{x-3+iy}\right|=2$

$\Rightarrow \sqrt{{\left(x-2\right)}^2+y^2}=2\sqrt{{\left(x-3\right)}^2+y^2}$

$\Rightarrow x^2+y^2-4x+4=4x^2+4y^2-24x+36$

$\Rightarrow 3x^2+3y^2-20x+32=0$

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

So, $\left(\alpha ,\beta \right)\equiv \left(\frac{10}{3},0\right)$

And,

$\gamma =\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$

Therefore,

$3\left(\alpha +\beta +\gamma \right)=3\left(\frac{10}{3}+\frac{2}{3}\right)=12$