Seja $z = (x \ , \ y)$, temos:\begin{matrix} |z-3i|^2 &=& x^2 + (y-3)^2 = [x^2 + (y-3)^2 \ , \ 0] &=& (9,0) \end{matrix} \begin{matrix} x^2 + (y-3)^2 = 9 & \color{royalblue}{(1)} \end{matrix}Continuando,\begin{matrix} |z+i|^2 &=& x^2 + (y+1)^2 &=& [x^2 + (y+1)^2 \ , \ 0] \\ \\ |z-2-i|^2 &=& (x-2)^2 + (y-1)^2 &=& [(x-2)^2+ (y-1)^2 \ , \ 0] \end{matrix} Segundo enunciado, \begin{matrix} [x^2 + (y+1)^2 \ , \ 0] &=& [(x-2)^2+ (y-1)^2 \ , \ 0] \\ \\ x^2 + (y+1)^2 &=& (x-2)^2+ (y-1)^2 \end{matrix} \begin{matrix} x+ y = 1 & \color{royalblue}{(2)} \end{matrix}Substituindo $(2)$ em $(1)$, encontramos, \begin{matrix} x_1 = -1 + \dfrac{\sqrt{14}}{2} &e& x_2 = 1 - \dfrac{\sqrt{14}}{2} \\ \\ y_1 = 2 - \dfrac{\sqrt{14}}{2} &e& y_2 = 2 + \dfrac{\sqrt{14}}{2} \end{matrix}Assim, os dois complexos, $z_1$ e $z_2$, são: \begin{matrix}z_1 = \left(-1 + \dfrac{\sqrt{14}}{2} \ \ , \ \ 2 - \dfrac{\sqrt{14}}{2}\right) &e& z_2 = \left(-1 - \dfrac{\sqrt{14}}{2} \ \ , \ \ 2 + \dfrac{\sqrt{14}}{2}\right) \end{matrix}Fazendo $z_1\cdot z_2$ temos o produto de todos os elementos de $S$: \begin{matrix} z_1\cdot z_2 = -3 + 3i \\ \\ Letra \ (D) \end{matrix}