Gọi $z=a+bi,(a,b\in R)\Rightarrow \overline{z}=a-bi$

Theo giả thiết:

$\left\{ \begin{align} & \left| z \right|=\sqrt{13} \\ & \left| z+2-i \right|=\sqrt{2}\left| \overline{z}+1-i \right| \\ \end{align} \right.$ $\Leftrightarrow \left\{ \begin{align} & \left| z \right|=\sqrt{13} \\ & \left| (a+2)+(b-1)i \right|=\sqrt{2}\left| (a+1)-(b+1)i \right| \\ \end{align} \right.$

$\left\{ \begin{align} & \sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{13} \\ & \sqrt{{{(a+2)}^{2}}+{{(b-1)}^{2}}}=\sqrt{2}.\sqrt{{{(a+2)}^{2}}+{{(b+1)}^{2}}} \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{matrix} {{a}^{2}}=9  \\ b=-2  \\ \end{matrix} \right.$$\Leftrightarrow \left\{ \begin{matrix} a=\pm 3  \\ b=-2  \\ \end{matrix} \right.$

Vậy $z=-3-2i$ hoặc $z=3-2i$

Đáp án đúng là C.