$(1-2i)z-\frac{2-i}{1+i}=(3-i)z$
$\Rightarrow (1-2i-3+i)z=\frac{2-i}{1+i}$
$\Rightarrow -(i+2)z=\frac{2-i}{1+i}$
$\Rightarrow z=\frac{i-2}{(1+i)(i+2)}$
$\Rightarrow z=\frac{i-2}{3i+1}$
$\Rightarrow z=\frac{(i-2)(3i-1)}{(3i+1)(3i-1)}$
$\Rightarrow z=\frac{7i+1}{10}$
$\Rightarrow z=\left ( \frac1{10},\frac7{10} \right )$