10. As shown in the figure, the volume of tetrahedron $D A B C$ is $\frac{1}{6}$, and it satisfies $\angle A C B=45^{\circ}, A D+B C+\frac{A C}{\sqrt{2}}=3$. Then $C D=$ $\qquad$
In the base $\triangle ABC$, construct $AH \perp BC$, then $AH=\frac{AC}{\sqrt{2}}$.
Thus $\frac{1}{6}=V_{DABC} \leqslant \frac{1}{6} AD \cdot BC \cdot \frac{AC}{\sqrt{2}}$
$\Rightarrow AD \cdot BC \cdot \frac{AC}{\sqrt{2}} \geqslant 1$, equality holds when $AD \perp$ plane $ABC$.
Also, $3=AD+BC+\frac{AC}{\sqrt{2}} \geqslant 3 \sqrt[3]{AD \cdot BC \cdot \frac{AC}{\sqrt{2}}} \geqslant 3 \Rightarrow AD=BC=\frac{AC}{\sqrt{2}}=1$,
so $AC=\sqrt{2}, AD=1$ and $AD \perp$ plane $ABC \Rightarrow CD=\sqrt{3}$.