3. Given complex numbers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=1, \\
b^{2}+b c+c^{2}=-1, \\
c^{2}+c a+a^{2}=\mathrm{i} .
\end{array}\right.
$$

Then $a b+b c+c a=$ $\qquad$
3. $\pm$ i.

From the problem, we have
$$
\begin{aligned}
a^{3}-b^{3} & =a-b, \\
b^{3}-c^{3} & =c-b, \\
c^{3}-a^{3} & =\mathrm{i}(c-a) .
\end{aligned}
$$

Adding the three equations, we get
$$
b=\frac{1+\mathrm{i}}{2} c+\frac{1-\mathrm{i}}{2} a \text {. }
$$

Substituting this into equation (1), we have
$$
\left(\frac{3}{2}-\mathrm{i}\right) a^{2}+\frac{3+\mathrm{i}}{2} a c+\frac{\mathrm{i}}{2} c^{2}=1 \text {. }
$$

Combining this with equation (3), we find that $a$ and $c$ are both non-zero and
$$
c=\frac{1}{a}-(1-\mathrm{i}) a \text {. }
$$

Thus, $a^{4}-\mathrm{i}(1-\mathrm{i}) a^{2}+\mathrm{i}=0$
$$
\Rightarrow a^{2}=1 \text { or } a^{2}=\mathrm{i} \text {. }
$$

If $a=1$, then $c=\mathrm{i}, b=0$, and the desired value is $\mathrm{i}$;
If $a=-1$, then $c=-\mathrm{i}, b=0$, and the desired value is i;
If $a^{2}=\mathrm{i}$, then $c^{2}+a c=0$.
Since $c \neq 0$, we have
$$
c=-a \Rightarrow\{a, c\}=\left\{\mathrm{e}^{\frac{\pi}{4} \mathrm{i}},-\mathrm{e}^{\frac{\pi}{4} \mathrm{i}}\right\} \text {. }
$$

Substituting into (1)-(2), we get $b=\frac{1}{a}$, and the desired value is $-\mathrm{i}$.
In summary, the desired values are $\pm \mathrm{i}$.