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(D) Let $z = x + iy$. The given equation is $\left|\frac{x+iy-2}{x+iy-3}\right|=2$.Squaring both sides,we get $\frac{(x-2)^2+y^2}{(x-3)^2+y^2}=4$.$(x-

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$12$

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(D) Let $z = x + iy$. The given equation is $\left|\frac{x+iy-2}{x+iy-3}\right|=2$.Squaring both sides,we get $\frac{(x-2)^2+y^2}{(x-3)^2+y^2}=4$.$(x-2)^2+y^2 = 4((x-3)^2+y^2)$.$x^2-4x+4+y^2 = 4(x^2-6x+9+y^2)$.$x^2-4x+4+y^2 = 4x^2-24x+36+4y^2$.$3x^2+3y^2-20x+32=0$.Dividing by $3$,we get $x^2+y^2-\frac{20}{3}x+\frac{32}{3}=0$.Comparing with the standard form $x^2+y^2+2gx+2fy+c=0$,we have $g=-\frac{10}{3}$ and $f=0$.The center $(\alpha, \beta) = (-g, -f) = \left(\frac{10}{3}, 0\right)$.The radius $\gamma = \sqrt{g^2+f^2-c} = \sqrt{\left(\frac{10}{3}\right)^2 - \frac{32}{3}} = \sqrt{\frac{100}{9} - \frac{96}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.Thus,$3(\alpha+\beta+\gamma) = 3\left(\frac{10}{3} + 0 + \frac{2}{3}\right) = 3\left(\frac{12}{3}\right) = 12$.

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

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