\(\left( {1 + i} \right)z + 2 - 3i = \left( {2 - i} \right)\left( {3 - 2i} \right) \Leftrightarrow z = \frac{{\left( {2 - i} \right)\left( {3 - 2i} \right) - 2 + 3i}}{{1 + i}}\)
=\(\frac{{2 - 4i}}{{1 + i}} = \frac{{\left( {2 - 4i} \right)\left( {1 - i} \right)}}{{{1^2} + {1^2}}} = \frac{{ - 2 - 6i}}{2} = - 1 - 3i\) \( \Rightarrow \left| z \right| = \sqrt {{1^2} + {3^2}} = \sqrt {10} .\)
