3. If $a+\frac{1}{b}=1, b+\frac{2}{c}=1$, then, $c+\frac{2}{a}$ equals ().
(A) 1
(B) 2
(C) 3
(D) 4
3. (B).

When $a=0$, $b=1$ contradicts $b+\frac{2}{c}=1$.
Therefore, $a \neq 0$ and $b \neq 1$.
From $a=\frac{b-1}{b}$, we get $\frac{2}{a}=\frac{2b}{b-1}$;
From $\frac{2}{c}=1-b$, we get $c=\frac{2}{1-b}$.
Thus, $c+\frac{2}{a}=2$.