(12) The complex number $z$ satisfies $|z|(3 z+2 \mathrm{i})=2(\mathrm{i} z-6)$, then $|z|=$
(12) 2 Hint: Direct calculation shows that
$$
|3 z+2 \mathrm{i}|^{2}-|\mathrm{i} z-6|^{2}=8\left(|z|^{2}-4\right) .
$$

From this, if $|z|>2$, then $|3 z+2 \mathrm{i}|>|\mathrm{i} z-6|$, and thus
$$
|| z|(3 z+2 \mathrm{i})|>|z(\mathrm{i} z-6)|,
$$

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

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

Simplifying, we get $r^{4}=16$, thus $r=2$, i.e., $|z|=2$.