3. Given real numbers $a, b, c$ satisfy
$$
a^{2}+b^{2}=4, b^{2}+c^{2}=8 \text {. }
$$

Then the minimum value of $a b+b c+\frac{\sqrt{2}}{2} a c$ is $\qquad$
3. $-3 \sqrt{2}$.

Notice that
$$
\begin{aligned}
(a+ & \sqrt{2} b+c)^{2} \\
& =a^{2}+2 b^{2}+c^{2}+2(\sqrt{2} a b+\sqrt{2} b c+a c) \\
& =\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+2 \sqrt{2}\left(a b+b c+\frac{\sqrt{2}}{2} a c\right) .
\end{aligned}
$$
$$
\begin{array}{l}
\text { Hence } a b+b c+\frac{\sqrt{2}}{2} a c \\
=\frac{\sqrt{2}}{4}(a+\sqrt{2} b+c)^{2}-\frac{\sqrt{2}}{4}(4+8) \\
=\frac{\sqrt{2}}{4}(a+\sqrt{2} b+c)^{2}-3 \sqrt{2} \geqslant-3 \sqrt{2} .
\end{array}
$$

When and only when $a+\sqrt{2} b+c=0$, the equality holds.
Therefore, the minimum value sought is $-3 \sqrt{2}$.