3. Given the complex number $z$ satisfies $3 z^{6}+2 \mathrm{i} z^{5}-2 z-3 \mathrm{i}=$ 0 . Then the modulus of $z$ ( ).
(A) greater than 1
(B) equal to 1
(C) less than 1
(D) cannot be determined
3. (B).

The original equation can be transformed into $z^{5}=\frac{2 z+3 \mathrm{i}}{3 z+2 \mathrm{i}}$.
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$, then
$$
\begin{aligned}
\left|z^{5}\right| & =\left|\frac{2 a+(2 b+3) \mathrm{i}}{3 a+(3 b+2) \mathrm{i}}\right|=\sqrt{\frac{4 a^{2}+(2 b+3)^{2}}{9 a^{2}+(3 b+2)^{2}}} \\
& =\sqrt{\frac{4\left(a^{2}+b^{2}\right)+12 b+9}{9\left(a^{2}+b^{2}\right)+12 b+4}} .
\end{aligned}
$$

If $a^{2}+b^{2}>1$, then the left side of equation (1) $=|z|^{5}>1$, while the right side $1$, so equation (1) does not hold.
Therefore, only $a^{2}+b^{2}=1$, i.e., $|z|=1$.