Let $$S= z \left\{\in \mathbb{C}:|\frac{z-6i}{z-2i}|=1\text{ and }|\frac{z-8+2i}{z+2i}|=\frac{3}{5} \right\}$$. Then $$\sum_{z\in s}^{}|z|^{2}$$ is equal to

We need to find $$\sum_{z \in S} |z|^2$$ where $$S = \left\{z \in \mathbb{C} : \left|\dfrac{z - 6i}{z - 2i}\right| = 1 \text{ and } \left|\dfrac{z - 8 + 2i}{z + 2i}\right| = \dfrac{3}{5}\right\}$$.

First, note that $$\left|\dfrac{z - 6i}{z - 2i}\right| = 1$$ implies $$|z - 6i| = |z - 2i|$$, so $$z$$ is equidistant from $$6i$$ and $$2i$$. The midpoint of these points is $$(0,4)$$, and the perpendicular bisector is the horizontal line $$y = 4$$, giving $$z = x + 4i$$ for some real $$x$$.

Next, substituting $$z = x + 4i$$ into the second condition yields $$z - 8 + 2i = (x - 8) + 6i$$ so $$|z - 8 + 2i| = \sqrt{(x-8)^2 + 36}$$, and $$z + 2i = x + 6i$$ so $$|z + 2i| = \sqrt{x^2 + 36}$$. Therefore,

$$\frac{\sqrt{(x-8)^2 + 36}}{\sqrt{x^2 + 36}} = \frac{3}{5}$$

Squaring both sides gives

$$\frac{(x-8)^2 + 36}{x^2 + 36} = \frac{9}{25}$$

and hence

$$25\bigl[(x-8)^2 + 36\bigr] = 9\bigl[x^2 + 36\bigr]$$

which expands to

$$25(x^2 - 16x + 64) + 900 = 9x^2 + 324$$

and simplifies to

$$25x^2 - 400x + 1600 + 900 = 9x^2 + 324$$

so that

$$16x^2 - 400x + 2176 = 0$$

or equivalently

$$2x^2 - 50x + 272 = 0$$

and thus

$$x^2 - 25x + 136 = 0$$

Solving this quadratic gives

$$x = \frac{25 \pm \sqrt{625 - 544}}{2} = \frac{25 \pm \sqrt{81}}{2} = \frac{25 \pm 9}{2}$$

Hence $$x = 17$$ or $$x = 8$$, which yields the points $$z_1 = 17 + 4i$$ and $$z_2 = 8 + 4i$$.

Finally, computing the squares of their moduli gives

$$|z_1|^2 = 17^2 + 4^2 = 289 + 16 = 305$$

$$|z_2|^2 = 8^2 + 4^2 = 64 + 16 = 80$$

so that

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

The answer is $$\boxed{385}$$ which corresponds to Option 1.