6. If the complex number $z$ satisfies $3 z^{6}+2 i \cdot z^{5}-2 z-3 i=$ 0 . Then $|z|=$ $\qquad$ .
6. $|z|=1$.
The equation is $z^{5}=\frac{2 z+3 i}{3 z+2 i}$. Let $z=a+b i$.
$$
\left|z^{5}\right|=\frac{|2 z+3 i|}{|3 z+2 i|}=\sqrt{\frac{4\left(a^{2}+b^{2}\right)+12 b+9}{9\left(a^{2}+b^{2}\right)+12 b+4}} \text {. }
$$

If $a^{2}+b^{2}>1$, then the left side of (1) $=|z|^{5}=\left(\sqrt{a^{2}+b^{2}}\right)^{5}>$ 1. But the right side has both the numerator and denominator positive, and the difference between the denominator and the numerator is $5\left(a^{2}+\right.$ $\left.b^{2}\right)-5>0$, making the fraction value less than 1. The equation cannot hold.
If $a^{2}+b^{2}<0$, similarly, it can be deduced that the equation does not hold.
Only when $a^{2}+b^{2}=1,|z|=1$.