Explanation

(a)
Given, $\left| \frac{z_{1} - 3z_{2}}{3 - z_{1}\overline{z_{2}}} \right| = 1,\ \ \ \left| z_{1} \right| \neq 3$
$\Rightarrow \ \ \ \left| z_{1} - 3z_{2} \right| = \left| 3 - z_{1}\overline{z_{2}} \right|\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack \because\ \left| \frac{z_{1}}{z_{2}} \right| = \frac{|z_{1}|}{|z_{2}|} \right\rbrack$
$\Rightarrow \ \ \ \ \left| z_{1} - 3z_{2} \right|^{2} = \left| 3 - z_{1}\overline{z_{2}} \right|^{2}$
$\Rightarrow \ \ \ \ \ \left( z_{1} - 3z_{2} \right)\left( \overline{z_{1}} - 3\overline{z_{2}} \right) = \left( 3 - z_{1}\overline{z_{2}} \right)\left( 3 - \overline{z_{1}}z_{2} \right)\ \ \ \ \ \lbrack\because\ \ \overline{\overline{z_{2}}} = z_{2}\rbrack$
$\Rightarrow \ \ \ \ \left| z_{1} \right|^{2} - 3z_{1}\overline{z_{2}} - 3z_{2}\overline{z_{1}} + 9\left| z_{2} \right|^{2}$
$= 9 - 3\overline{z_{1}}z_{2} - 3z_{1}\overline{z_{2}} + \left| z_{1} \right|^{2}\left| z_{2} \right|^{2}$
$\Rightarrow \ \ \ \ \left| z_{1} \right|^{2} + 9\left| z_{2} \right|^{2} - 9 - \left| z_{1} \right|^{2}\left| z_{2} \right|^{2} = 0$
$\Rightarrow \ \ \ \ \left( 9 - \left| z_{1} \right|^{2} \right)\left( 1 - \left| z_{2} \right|^{2} \right) = 0$
$\Rightarrow \ \ \ \left| z_{1} \right|^{2} = 9$ or $\left| z_{2} \right|^{2} = 1$
$\Rightarrow \ \ \ \ \left| z_{1} \right| = 3$ or $\left| z_{2} \right| = 1$
$\therefore\ \ \ \ \left| z_{2} \right| = 1$ [but $|z_{1}| \neq 3$ given]