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 $C D=$ $\qquad$ .
Since $\frac{1}{3} A D \cdot\left(\frac{1}{2} \cdot B C \cdot A C \cdot \sin 45^{\circ}\right) \geqslant V_{D A B C}=\frac{1}{6}$, that is $A D \cdot B C \cdot \frac{A C}{\sqrt{2}} \geqslant 1$.

Also, $3=A D+B C+\frac{A C}{\sqrt{2}} \geqslant 3 \cdot \sqrt[3]{A D \cdot B C \cdot \frac{A C}{\sqrt{2}}} \geqslant 3$,
the equality holds if and only if $A D=B C=\frac{A C}{\sqrt{2}}=1$, at this time $A B=1, A D \perp$ plane $A B C$, so $D C=\sqrt{3}$.