Given $|z-2|=|z-1|$

On squaring both sides, we get

$\left(z-2\right)\left(\overline{z}-2\right)=\left(z-1\right)\left(\overline{z}-1\right)$

$\Rightarrow z\overline{z}-2\overline{z}-2z+4=z\overline{z}-\overline{z}-z+1$

$\Rightarrow \overline{z}+z=3$

Now, putting $z=x+iy$, we get

$\left(x+iy\right)+\left(x-iy\right)=3$

$\Rightarrow x=\frac{3}{2}$

Hence, locus of $z$ is a straight line parallel to $y-$axis