(A) 413

(B) 398

(C) 385

(D) 423

Correct Answer: 385

Explanation

Let $z = x + iy$.

First condition:

$$ \left|\frac{z-6i}{z-2i}\right|=1 \;\Rightarrow\; |z-6i|=|z-2i| $$

This represents the perpendicular bisector of the points $2i$ and $6i$.

Midpoint $=4i$ and the line is horizontal:

$$ y=4 $$

Second condition:

$$ \left|\frac{z-(8-2i)}{z+2i}\right|=\frac{3}{5} \Rightarrow |z-(8-2i)|=\frac{3}{5}|z+2i| $$

Squaring and simplifying:

$$ 25\big[(x-8)^2+(y+2)^2\big]=9\big[x^2+(y+2)^2\big] $$

$$ x^2+y^2-25x+4y+104=0 $$

Completing squares:

$$ (x-12.5)^2+(y+2)^2=56.25 $$

This is a circle with center $(12.5,-2)$ and radius $7.5$.

Intersection with $y=4$:

$$ (x-12.5)^2+36=56.25 $$

$$ (x-12.5)^2=20.25 \Rightarrow x=17,\;8 $$

Hence,

$$ z_1=8+4i,\quad z_2=17+4i $$

Required sum:

$$ |8+4i|^2=8^2+4^2=80 $$

$$ |17+4i|^2=17^2+4^2=305 $$

$$ \sum |z|^2=80+305=\boxed{385} $$

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