Let $$z$$ be a complex number such that the real part of $$\frac{z-2i}{z+2i}$$ is zero. Then, the maximum value of $$|z - (6 + 8i)|$$ is equal to

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

Real part = 0 means: $$\text{Re}\left(\frac{(x+i(y-2))(x-i(y+2))}{|z+2i|^2}\right) = 0$$.

Numerator real part: $$x^2 + (y-2)(y+2) = x^2 + y^2 - 4 = 0$$.

So $$|z|^2 = 4$$, i.e., z lies on circle $$x^2+y^2=4$$.

Maximum of $$|z-(6+8i)|$$ on this circle = distance from centre to (6,8) + radius = $$\sqrt{36+64}+2 = 10+2 = 12$$.

The correct answer is Option 1: 12.