$$\begin{aligned} &\text { Let } z=x+i y\\ &\begin{aligned} & |z-2-i|=3 \Rightarrow(x-2)^2+(y-1)^2=3^2 \\ & \operatorname{Re}(z-i z)=\operatorname{Re}(x+i y-i x+y)=x+y \Rightarrow x+y=2 \\ & \Rightarrow A=\left\{(x, y):(x-2)^2+(y-1)^2=3^2, x, y \in R\right\}, \\ & B=\{(x, y): x+y=2\} \\ & \Rightarrow x-2=-y \Rightarrow y^2+(y-1)^2=3^2 \\ & \Rightarrow 2 y^2-2 y-8=0 \Rightarrow y^2-y-4=0 \\ & y_1+y_2=1, y_1 y_2=-4 \\ & \Rightarrow y_1^2+y_2^2 \\ & =\left(y_1+y_2\right)^2-2 y_1 y_2=9 \\ & \Rightarrow x_1+x_2=4\left(y_1+y_2\right)=3, \\ & x_1 x_2=\left(2-y_1\right)\left(2-y_2\right)=4-2\left(y_1+y_2\right)+y_1 y_2=-2 \\ & \Rightarrow x_1^2+x_2^2=\left(x_1+x_2\right)^2-2 x_1 x_2=13 \\ & \because S=\left\{\left(x_1, y_1\right),\left(x_2, y_2\right)\right\} \\ & \Rightarrow \sum_{z \in S}|z|^2=\left(x_1^2+y_1^2\right)+\left(x_2^2+y_2^2\right)=22 \end{aligned} \end{aligned}$$