$$
\begin{array}{l}
\text { 3. As shown in figure } 1, \angle X O Y \\
=90^{\circ}, \quad O W \text { bisects } \\
\angle X O Y, P A \perp O X, P B \\
\perp O Y, P C \perp O W \text {. If } O A \\
+O B+O C=1 \text {, then } O C \\
=(\quad) .
\end{array}
$$
3. As shown in figure $1, \angle X O Y$
(A) $2-\sqrt{2}$
(B) $\sqrt{2}-1$
(C) $\sqrt{6}-2$
(D) $2 \sqrt{3}-3$
$$
3. B.

Extend $C P$ to intersect $O Y$ at point $D$, it is easy to know that $B D=P B=O A$.
Then $O A+O B=O B+B D=O D=\sqrt{2} O C$. Therefore, $1=O A+O B+O C=(\sqrt{2}+1) O C$, which means $O C=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$.