$\left|z\omega \right|=1$

If $\left|z\right|=r$, then $\left|\omega \right|=\frac{1}{r}$

Let $\arg \left(z\right)=\theta$

$\therefore \arg \left(\omega \right)=\left(\theta -\frac{3\pi }{2}\right)$

So,

$z=r{e}^{i\theta }$

$\Rightarrow \overline{z}=r{e}^{i\left(-\theta \right)}$
$\Rightarrow \omega =\frac{1}{r}{e}^{i\left(\theta -\frac{3\pi }{2}\right)}$

Now, consider
$\frac{1-2\overline{z}\omega }{1+3\overline{z}\omega }=\frac{1-2e^{i\left(-{\displaystyle \frac{3\pi }{2}}\right)}}{1+3e^{i\left(-{\displaystyle \frac{3\pi }{2}}\right)}}=\left(\frac{1-2i}{1+3i}\right)$

$=\frac{\left(1-2i\right)\left(1-3i\right)}{\left(1+3i\right)\left(1-3i\right)}=\frac{1-5i+6i^2}{10}=\frac{-5+5i}{10}=-\frac{1}{2}\left(1+i\right)$

Then,

$\mathrm{principal} \arg \left(\frac{1-2\overline{\mathrm{z}}\omega }{1+3\overline{\mathrm{z}}\omega }\right)$

$=\mathrm{principal} \arg \left(-\frac{1}{2}\left(1+i\right)\right)$

$=-\left(\pi -\frac{\pi }{4}\right)=\frac{-3\pi }{4}$