We have,
\(\arg \left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\)
\(\begin{array}{lc}
\Rightarrow & \arg (z-1)-\arg (z-3 i)=\frac{\pi}{2} \\
\Rightarrow & \arg [(x-1)+i y]-\arg [x+(y-3) i]=\frac{\pi}{2} \\
\Rightarrow & \tan ^{-1} \frac{y}{x-1}-\tan ^{-1} \frac{y-3}{x}=\frac{\pi}{2} \\
\Rightarrow & \tan ^{-1}\left[\frac{y}{1+\frac{y}{x-1} \cdot \frac{y-3}{x}}\right]=\frac{\pi}{2} \\
\Rightarrow & \frac{x y-(x-1)(x-3)}{x(x-1)+y(x-3)}=\tan \frac{\pi}{2} \\
\Rightarrow & \frac{x y-(x-1)(y-3)}{x(x-1)+y(y-3)}=\frac{1}{0} \\
\Rightarrow & x(x-1)+y(y-3)=0 \\
\Rightarrow & x y=0
\end{array}\)
\(\begin{aligned}
& \Rightarrow \quad\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{1}{4}+\frac{9}{4} \\
& \Rightarrow \quad\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{\sqrt{10}}{2}\right)^2
\end{aligned}\)
Which is a circle with centre \(\left(\frac{1}{2}, \frac{3}{2}\right)\) and radius \(\frac{\sqrt{10}}{2}\).
\(\therefore \quad z \in C:(3-i) z+(3+i) \bar{z}-6 > 0,\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\)