Let $z=x+i y$
$\therefore|z|^{2}+|z-3|^{2}+|z-i|^{2}$
$$
\begin{aligned}
=|x+i y|^{2}+\mid &\mid x-3)+\left.i y\right|^{2} \\
&+|x+i(y-1)|^{2}
\end{aligned}
$$
$=x^{2}+y^{2}+(x-3)^{2}+y^{2}+x^{2}+(y-1)^{2}$
$=x^{2}+y^{2}+x^{2}-6 x+9+y^{2}+x^{2}+y^{2}$
$=3 x^{2}+3 y^{2}-6 x-2 y+10$
$=3\left(x^{2}-2 x+1\right)+3\left(y^{2}-\frac{2}{3} y+\frac{1}{9}\right)$
$\quad=3(x-1)^{2}+3\left(y-\frac{1}{3}\right)^{2}+\frac{20}{3}$
It is minimum, when $x-1=0$ and $y-\frac{1}{3}=0$
$\therefore \quad x=1 \quad$ and $\quad y=\frac{1}{3}$
$\therefore \quad z=1+\frac{1}{3} i$