$\text{ Suppose, phases of $z_1,z_2$ and $z_3$ are $\theta_1, \theta_2$ and $\theta_3$ respectively}$
$\text{and given: $|z_1| = |z_2|= |z_3| = r$}$
$\text{$\therefore z_1 = r e^{i\theta_1}, z_2 = r e^{i\theta_2}$ and $z_2 = r e^{i\theta_3}$ }$
$\text{$\arg (\frac{z_1}{z_2}) = \arg(\frac{re^{i\theta_1}}{re^{i\theta_2}}) = \arg(e^{i(\theta_1 – \theta_2)})= \theta_1 – \theta_2$ }$
$\text{$\arg(\frac{z_1-z_3}{z_2 – z_3}) = \arg \left(\frac{\frac{z_1}{z_3} – 1}{\frac{z_2}{z_3} – 1}\right) = \arg \left( \frac{e^{i(\theta_1 – \theta_3) } – 1}{e^{i(\theta_2 – \theta_3)} – 1} \right) = \arg \left( \frac{(\cos(\theta_1 – \theta_3)-1) + i\sin(\theta_1 – \theta_3)}{(\cos(\theta_2 – \theta_3) – 1) + i\sin(\theta_2 – \theta_3)} \right) $}$
$\text{ $= \arg \left ( \frac{-2\sin^2(\frac{\theta_1-\theta_3}{2}) + i \sin (\theta_1 – \theta_3)}{-2\sin^2(\frac{\theta_2-\theta_3}{2}) + i \sin (\theta_2 – \theta_3)}\right) $}$
$\text{ $ = \arg (-2\sin^2(\frac{\theta_1-\theta_3}{2}) + i \sin (\theta_1 – \theta_3)) – \arg (-2\sin^2(\frac{\theta_2-\theta_3}{2}) + i \sin (\theta_2 – \theta_3)) $ }$
$\text{$\because -2\sin^2(\frac{\theta_{\{1,2\}}-\theta_3}{2}) < 0$ and $\sin (\theta_{\{1,2\}} – \theta_3) \geq 0$}$
$\text{$\therefore = \pi + \tan^{-1}\left(\frac{\sin (\theta_1 – \theta_3)}{-2\sin^2(\frac{\theta_1-\theta_3}{2})}\right) – \pi – \tan^{-1}\left(\frac{\sin (\theta_2 – \theta_3)}{-2\sin^2(\frac{\theta_2-\theta_3}{2})}\right) $}$
$\text{$= \tan^{-1} \left(\cot \frac{\theta_3 – \theta_1}{2} \right) – \tan^{-1} \left(\cot \frac{\theta_3 – \theta_2}{2} \right)$}$
$\text{$= \frac{\pi}{2} – \cot^{-1}\left(\cot \frac{\theta_3 – \theta_1}{2}\right) – \frac{\pi}{2} + \cot^{-1}\left(\cot \frac{\theta_3 – \theta_2}{2}\right)$}$
$\text{$= -(\frac{\theta_3 – \theta_1}{2}) + \frac{\theta_3 – \theta_2}{2} = \frac{\theta_1 – \theta_2}{2}$}$
$\text{Therefore, $\arg (\frac{z_1}{z_2}) = (\theta_1 – \theta_2) $ and $\arg(\frac{z_1-z_3}{z_2 – z_3}) = \frac{(\theta_1 – \theta_2)}{2}$}$
$\textbf{Answer: A }$