7. Given positive numbers $a, b, c$ satisfying $a^{2}+b^{2}+2 c^{2}=1$. Then the maximum value of $\sqrt{2} a b+2 b c+7 a c$ is $\qquad$
7. $2 \sqrt{2}$.
$$
\begin{array}{l}
\text { Given } \sqrt{2}(\sqrt{2} a b+2 b c+7 a c) \\
=a \cdot 2 b+2 \sqrt{2} b \cdot c+7 a \cdot \sqrt{2} c \\
\leqslant \frac{a^{2}+4 b^{2}}{2}+2 \times \frac{2 b^{2}+c^{2}}{2}+7 \times \frac{a^{2}+2 c^{2}}{2} \\
=4\left(a^{2}+b^{2}+2 c^{2}\right) \leqslant 4 \\
\Rightarrow \sqrt{2} a b+2 b c+7 a c \leqslant 2 \sqrt{2} .
\end{array}
$$

When and only when $a=2 b, c=\sqrt{2} b, a=\sqrt{2} c$, i.e., $a=$ $\frac{2}{3}, b=\frac{1}{3}, c=\frac{\sqrt{2}}{3}$, the equality holds in all the above expressions.
Therefore, the maximum value is $2 \sqrt{2}$.