6. The complex number $z$ satisfies
$$
|z|(3 z+2 \mathrm{i})=2(\mathrm{i} z-6) \text {. }
$$

Then $|z|$ equals $\qquad$ .
6. 2 .

Solution 1 Direct calculation shows that
$$
|3 z+2 \mathrm{i}|^{2}-|\mathrm{i} z-6|^{2}=8\left(|z|^{2}-4\right) \text {. }
$$

From this, if $|z|>2$, then
$$
\begin{array}{l}
|3 z+2 \mathrm{i}|>|\mathrm{i} z-6| \\
\Rightarrow|| z|(3 z+2 \mathrm{i})|>|2(\mathrm{i} z-6)|,
\end{array}
$$

which contradicts the given condition;
if $|z|<2$, then
$$
|| z \mid(3 z+2 \text { i) }|<| 2(z+6 \text { i) } \mid,
$$

which is also a contradiction.
Therefore, $|z|=2$.
Solution 2 Let $|z|=r$. Substituting into the condition, we get
$$
z=\frac{-(12+2 r \mathrm{i})}{3 r-2 \mathrm{i}} \text {. }
$$

Then $r^{2}=|z|^{2}=\frac{12^{2}+(2 r)^{2}}{(3 r)^{2}+(-2)^{2}}$.
Solving this, we get $r=2$, i.e., $|z|=2$.