jmc

Let $z=x+yi,$ where $x$ and $y$ are real numbers. Then $$\begin{array}{rl}|z-3{|}^2+|z-5+2i{|}^2+|z-1+i{|}^2& =|x+yi-3{|}^2+|x+yi-5+2i{|}^2+|x+yi-1+i{|}^2\\ & =|(x-3)+yi{|}^2+|(x-5)+(y+2)i{|}^2+|(x-1)+(y+1)i{|}^2\\ & =(x-3{)}^2+y^2+(x-5{)}^2+(y+2{)}^2+(x-1{)}^2+(y+1{)}^2\\ & =3x^2-18x+3y^2+6y+40\\ & =3(x-3{)}^2+3(y+1{)}^2+10\\ & \ge 10.\end{array}$$Equality occurs when $x=3$ and $y=-1,$ so the minimum value is $\boxed{10}.$