Đặt \(z = a + bi (a, b \in R)\).

Ta có: \(\overline z = a - bi\) và

\((2 - i)\overline z = (2 - i)(a - bi)= 2a - b - (2b + a)i\)

Do đó 

\(\begin{array}{*{20}{l}}
\begin{array}{l}
z + \left( {2 - i} \right)\bar z = 13 - 3i\\
 \Leftrightarrow a + bi + 2a - b - \left( {2b + a} \right)i = 13 - 3i
\end{array}\\
\begin{array}{l}
 \Leftrightarrow 3a - b - \left( {a + b} \right)i = 13 - 3i\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{3a - b = 13}\\
{a + b = 3}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{a = 4}\\
{b =  - 1}
\end{array}} \right.
\end{array}
\end{array}\)

Vậy \(\left| z \right| = \sqrt {{a^2} + {b^2}}  = \sqrt {17} \)