Cho số phức $z=a+bi\left( a,b\in \mathbb{R},a>0 \right)$ thỏa mãn...

Ta có $z.\overline{z}-12\left| z \right|+\left( z-\overline{z} \right)=13-10i\Leftrightarrow {{a}^{2}}+{{b}^{2}}-12\sqrt{{{a}^{2}}+{{b}^{2}}}+2bi=13-10i$
$\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}-12\sqrt{{{a}^{2}}+{{b}^{2}}}=13 \\
& 2b=-10 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+25-12\sqrt{{{a}^{2}}+25}=13 \\
& b=-5 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& \sqrt{{{a}^{2}}+25}=13 \\
& \sqrt{{{a}^{2}}+25}=-1\left( VN \right) \\
\end{aligned} \right. \\
& b=-5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\pm 12 \\
& b=-5 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=12 \\
& b=-5 \\
\end{aligned} \right., $ vì $ a>0$.
Vậy $S=a+b=7.$