Passing the Toll

Take $z=x+iy$ , then the expression becomes

$f=x^2+y^2+(x-3)^2+y^2+x^2+(y-6)^2$ $=2x^2+2y^2+(x-3)^2+(y-6)^2$

To find the minimum, we equate the partial derivatives of $f$ w.r,t $x$ and $y$ to zero.

$\frac{\partial f}{\partial x}=4x+2(x-3)=0 \implies x=1$

and

$\frac{\partial f}{\partial y}=4y+2(y-6)=0 \implies y=2$

Thus, the minimum occurs at $(x,y)=(1,2)$

Substituting for $(x,y)$ in $f$ , the minimum is 30. Hence, the required answer is $\frac{30}{5}=\boxed{6}$

Let $z=x+yi$ . Then

$\begin{aligned} \chi & = |z|^2 + |z-3|^2 + |z-6i|^2 \\ & = |x+yi|^2 + |x-3+yi|^2 + |x+(y-6)i|^2 \\ & = x^2+y^2 + (x-3)^2+y^2 + x^2+(y-6)^2 \\ & = 3x^2+3y^2 -6x - 12y + 45 \\ & = 3\left((x-1)^2+(y-2)^2+10\right) \end{aligned}$

Note that $\chi > 0$ and it is minimum when $x=1$ and $y=2$ . $\implies \chi_{max} = 30$ $\implies \dfrac \chi 5 = \boxed{6}$

A simple solution based on Centre of Mass.Convert the given complex nos into a Cartesian system.Now at the three vertices place 3 masses(equal).Clearly the given eq suggests us to find a sort of Moment of Inertia of the system about a given axis.And clearly it's minimum when we choose the axis to be passing through the centre of mass of the system.And clearly for three particles places at $(x1,y1),(x2,y2),(x3,y3)$ the centre of mass is $(x1+x2+x3/3,y1+y2+y3/3)$ ....And this is nothing but the Centroid of the Triangle.Hence here the Centroid is $(1,2)$ .Calculate the distances of the points $(0,0),(3,0),(0,6)$ .And $0.2*30=6$ ....... $Ans$ ...