Given, $\left|z–2\right|=\min \left\{\left|z–1\right|,\left|z–5\right|\right\} \cdots \left(\mathrm{i}\right)$ 

case $\left(1\right)$. when  $\left|\mathrm{z}-1\right|<\left|\mathrm{z}-5\right|$
i.e., $\left|\mathrm{z}-2\right|=\left|\mathrm{z}-1\right| \dots \left(\mathrm{ii}\right)$

so, here it is clear that $z$ lies on perpendicular bisector of line joining oints $\left(1,0\right),\left(2,0\right)$ on real axis.,
$\Rightarrow \mathrm{ }Re\left(z\right)=\frac{3}{2}$  which satisfy $\left|\mathrm{z}-1\right|<\left|\mathrm{z}-5\right|.$

again  case $\left(2\right)$  when  $\left|\mathrm{z}-5\right|<\left|\mathrm{z}-1\right|$
$\Rightarrow$$\left|\mathrm{z}-2\right|=\left|\mathrm{z}-5\right|$ 

here$z$ lies on real line which perpendicular bisector of line joinig points $\left(2,0\right),\left(5,0\right)$
$\Rightarrow \mathrm{ }Re\left(z\right)=\frac{7}{2}$ which satisfy $\left|\mathrm{z}-5\right|<\left|\mathrm{z}-1\right|$ .

Hence option $3$ is correct.