3. Given the complex number $z$ satisfies $z^{2}+2 z=\bar{z} \neq z$ (where $\bar{z}$ denotes the conjugate of $z$), then the product of all possible values of $z$ is $\qquad$ .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R})$, then $z^{2}+2 z=\bar{z} \Rightarrow a^{2}-b^{2}+2 a+(2 a b+2 b) \mathrm{i}=a-b \mathrm{i}$ $\Rightarrow\left\{\begin{array}{l}a^{2}-b^{2}+2 a=a, \\ 2 a b+2 b=-b\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}+a=b^{2}, \\ 2 a b+3 b=0 .\end{array}\right.\right.$
Since $\bar{z} \neq z \Rightarrow b \neq 0 \Rightarrow a=-\frac{3}{2} \Rightarrow b^{2}=a^{2}+a=\frac{3}{4} \Rightarrow b= \pm \frac{\sqrt{3}}{2}$,
thus $z=-\frac{3}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$. Therefore, the product of all possible values of $z$ is $|z|^{2}=3$.