Official solution
As \[\left| {z\omega } \right| = 1\]
\[\Rightarrow\] If \[\left| z \right| = r\], then \[\left| \omega \right| = {1 \over r}\]
Let \[\arg (z) = \theta \]
\[\therefore\] \[\arg (\omega ) = \left( {\theta - {{3\pi } \over 2}} \right)\]
So, \[z = r{e^{i\theta }}\]
\[ \Rightarrow \overline z = r{e^{i\theta }}\]
\[\omega = {1 \over r}{e^{i\left( {\theta - {{3\pi } \over 2}} \right)}}\]
Now, consider
\[{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }} = {{1 - 2{e^{i\left( { - {{3\pi } \over 2}} \right)}}} \over {1 + 3{e^{i\left( { - {{3\pi } \over 2}} \right)}}}} = \left( {{{1 - 2i} \over {1 + 3i}}} \right)\]
\[ = {{(1 - 2i)(1 - 3i)} \over {(1 + 3i)(1 - 3i)}} = - {1 \over 2}(1 + i)\]
\[\therefore\] \[prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)\]
\[ = prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)\]
\[ = \left( { - {1 \over 2}(1 + i)} \right)\]
\[ = - \left( {\pi - {\pi \over 4}} \right) = {{ - 3\pi } \over 4}\]
So, option (2) is correct.