$$\begin{aligned} & n=\frac{z-2 i}{z+2 i} \\ & \text { Let } z=x+i y \\ & n=\frac{x+(y-2) i}{x+(y+2) i} \times\left(\frac{x-(y+2) i}{x-(y+2) i}\right) \\ & \operatorname{Re}(n)=\frac{x^2+(y-2)(y+2)}{x^2+(y+2)^2}=0 \\ & \Rightarrow x^2+(y-2)(y+2)=0 \end{aligned}$$

$$\begin{aligned} & \Rightarrow x^2+y^2-4=0 \\ & \Rightarrow x^2+y^2=4 \\ & \text { also, }|z-(6+8 i)| \leq|z|+|-6-8 i| \\ & |z-(6+8 i)| \leq 2+10=12 \end{aligned}$$

Hence, Maximum value of $$|z-(6+8 i)|$$ is 12.