3. Given $a b c=1, a+b+c=2, a^{2}+b^{2}+c^{2}=3$. Then the value of $\frac{1}{a b+c-1}+\frac{1}{b c+a-1}+\frac{1}{c a+b-1}$ is ( ).
(A) 1
(B) $-\frac{1}{2}$
(C) 2
(D) $-\frac{2}{3}$
3. D
3. $a b+c-1=(a-1)(b-1), b c+a-i=(b-1)$
- $(c-1), c a+b-1=(c-1)(a-1)$ Simplify and solve.