Solution
Let $\mathrm{z}=\mathrm{x}+\mathrm{iy} ; \mathrm{A}:|\mathrm{z}-2-\mathrm{i}|=3 \Rightarrow|(\mathrm{x}-2)+(\mathrm{y}-1) \mathrm{i}|=3 \Rightarrow(\mathrm{x}-2)^{2}+(\mathrm{y}-1)^{2}=9$
\[
\begin{aligned}
& B=\operatorname{Re}(z-i z)=2 \Rightarrow \operatorname{Re}((x+y)+i(y-x))=2 \Rightarrow x+y=2 \Rightarrow x=2-y \ldots(2) ; \text { On solving (1) and (2) } \\
& (2-y-2)^{2}+(y-1)^{2}=9 \Rightarrow 2 y^{2}-2 y-8=0 \Rightarrow y^{2}-y-4=0 \\
& y=\frac{1 \pm \sqrt{17}}{2}, x=2-y=\frac{3 \pm \sqrt{17}}{2} \Rightarrow z=\frac{3-\sqrt{17}}{2}+\left(\frac{1+\sqrt{17}}{2}\right) i ; z=\frac{3+\sqrt{17}}{2}+\left(\frac{1-\sqrt{17}}{2}\right) i \\
& \sum_{z \in S}|z|^{2}=\left(\left(\frac{3-\sqrt{17}}{2}\right)^{2}+\left(\frac{1+\sqrt{17}}{2}\right)^{2}\right)+\left(\left(\frac{3+\sqrt{17}}{2}\right)^{2}+\left(\frac{1-\sqrt{17}}{2}\right)^{2}\right)=\frac{2[9+17]}{4}+\frac{2[1+17]}{4}=13+9=22
\end{aligned}
\]
.
