JEE Main 20231 Feb, Morning shiftMathematicsMediumComplex Numbers and Quadratic Equations264× since 2002If the center and radius of the circle ∣z−2z−3∣=2\left| {{{z - 2} \over {z - 3}}} \right| = 2z−3z−2=2 are respectively (α,β)(\alpha,\beta)(α,β) and γ\gammaγ, then 3(α+β+γ)3(\alpha+\beta+\gamma)3(α+β+γ) is equal to :AOption A. 12BOption B. 10COption C. 11DOption D. 9CheckSkip · show solution
