If Z is complex number such that $\frac{z-i}{z-1}$ is purely imaginary then minimum value of $|z-(3+3i)|$ is 

$\frac{z-i}{z-1}=\frac{x+iy-i}{x+iy-1}=-\frac{x(x-1)+y(y-1)+i(-y-x+1)}{{(x-1)}^2+y^2}$

Put=0

$\Rightarrow x^2-x+y^2-y=0\Rightarrow {(x-\frac{1}{2})}^2+{(y-\frac{1}{2})}^2=\frac{1}{2}$

$C=(\frac{1}{2},\frac{1}{2})$

$|z-3+3i|=\left|PC\right|=\frac{1}{\sqrt{2}}$

$\le \frac{5}{\sqrt{2}}-\frac{1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$