$\begin{aligned} & \text { Given that }\left|\frac{z_1-3 z_2}{3-z_1 \bar{z}_2}\right|=1,\left|z_1\right| \neq 3 \\ & \Rightarrow\left|z_1-3 z_2\right|=\left|3-z_1 \bar{z}_2\right| \quad\left[\because\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\right] \\ & \Rightarrow \quad\left|z_1-3 z_2\right|^2=\left|3-z_1 \bar{z}_2\right|^2 \\ & \Rightarrow \quad\left(z_1-3 z_2\right)\left(\bar{z}_1-3 \bar{z}_2\right)=\left(3-z_1 \bar{z}_2\right)\left(3-\bar{z}_1 z_2\right) \\ & {\left[\because \bar{z}=z_2\right]} \\ & \Rightarrow \quad\left|z_1\right|^2-3 z_1 \bar{z}_2-3 z_2 \bar{z}_1+9\left|z_2\right|^2 \\ & =9-3 \bar{z}_1 z_2-3 z_1 \bar{z}_2+\left|z_1\right|^2\left|z_2\right|^2 \\ & \Rightarrow \quad\left|z_1\right|^2+9\left|z_2\right|^2-9-\left|z_1\right|^2\left|z_2\right|^2=0 \\ & \Rightarrow \quad\left(9-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)=0 \\ & \Rightarrow \quad\left|z_1\right|^2=9 \text { or }\left|z_2\right|^2=1 \\ & \Rightarrow \quad\left|z_1\right|=3 \mid \text { or }\left|z_2\right|=1 \\ & \end{aligned}$