We have,
$$
\begin{aligned}
& z_1=2-3 i, z_2=-1+i \\
& \text { and } \arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{2}
\end{aligned}
$$



$\therefore\left|z-z_1\right|^2+\left|z_2-z_2\right|^2=\left|z_1-z_2\right|^2$

$$
\begin{aligned}
& \Rightarrow|z-(2-3 i)|^2+|z-(-1+i)|^2=|2-3 i+1-i|^2 \\
& \Rightarrow(x-2)^2+(y+3)^2+(x+1)^2+(y-1)^2=9+16 \\
& {[\because z=x+i y]} \\
& \Rightarrow x^2-4 x+4+y^2+6 y+9+x^2+2 x+1+y^2 \\
& -2 y+1=25 \\
& \Rightarrow \quad 2 x^2+2 y^2-2 x+4 y-10=0 \\
& \Rightarrow \quad x^2+y^2-x+2 y-5=0 \\
&
\end{aligned}
$$
Now equation of line paring through $z_1(2-3)$ and $z_2(-1,1)$ is
$$
\begin{array}{ll}
y+3=\frac{1+3}{-1-2}(x-2) & \Rightarrow y+3=\frac{4}{-3}(x-2) \\
\Rightarrow-3 y-a=4 x-8 & \Rightarrow-3 y-9=4 x-8 \\
\Rightarrow 4 x+3 y+1=0 &
\end{array}
$$
According to given condition, $z$ should not lies on $z_1$ and $z_2$
$$
\therefore 4 x+3 y+1>0
$$