3. Let $a=\sqrt{2}-1, b=3-2 \sqrt{2}, c=\sqrt{3}-\sqrt{2}$. Then the size relationship of $a, b, c$ is ( ).
(A) $a>b>c$
(B) $c>b>a$
(C) $c>a>b$
(D) $a>c>b$
3. D.

Notice that $a-c=2 \sqrt{2}-(\sqrt{3}+1)$.
And $(2 \sqrt{2})^{2}=8>(\sqrt{3}+1)^{2}=4+2 \sqrt{3}$, so $a>c$.

Also, $c-b=\sqrt{2}+\sqrt{3}-3$, and $(\sqrt{2}+\sqrt{3})^{2}=5+2 \sqrt{6}>3^{2}=9$, so $c>b$.
Therefore, $a>c>b$.