6. Let the complex number $z$ satisfy $|z|=1$. Then
$$
|(z+1)+\mathrm{i}(7-z)|
$$

cannot be ( ).
(A) $4 \sqrt{2}$
(B) $4 \sqrt{3}$
(C) $5 \sqrt{2}$
(D) $5 \sqrt{3}$
6. D.

Notice,
$$
\begin{array}{l}
|(z+1)+\mathrm{i}(7-z)|=|1-\mathrm{i}||z-3+4 \mathrm{i}| \\
=\sqrt{2}|z-(3-4 \mathrm{i})| .
\end{array}
$$

Since $z$ is on the unit circle, the range of the above expression is $[4 \sqrt{2}, 6 \sqrt{2}]$.