3. Let the three sides of a right-angled triangle be $a$, $b$, and $c$ ($c$ being the hypotenuse), with the perimeter being 6. Then the minimum value of $c$ is $\qquad$
3. $6(\sqrt{2}-1)$.

Let the side $a$ of a right-angled triangle be opposite to angle $\alpha$. Then $0<\alpha<\frac{\pi}{2}$.
Hence $a=c \sin \alpha, b=c \cos \alpha$.
$$
\begin{array}{l}
\text { Also } a+b+c=6 \\
\Rightarrow c(\sin \alpha+\cos \alpha+1)=6 \\
\Rightarrow c\left[\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right)+1\right]=6 \\
\Rightarrow c=\frac{6}{\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right)+1} .
\end{array}
$$

Since $\sin \left(\alpha+\frac{\pi}{4}\right)$ reaches its maximum value of 1 when $\alpha=\frac{\pi}{4}$, therefore, when $\alpha=\frac{\pi}{4}$, $c$ reaches its minimum value $6(\sqrt{2}-1)$.