3. Given real numbers $a, b, c$ satisfy
$$
\left(-\frac{1}{a}+b\right)\left(\frac{1}{a}+c\right)+\frac{1}{4}(b-c)^{2}=0 \text {. }
$$

Then the value of the algebraic expression $a b+a c$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2
3. A.

The given equation can be rewritten as
$$
4(a b+1)(a c+1)+(a b-a c)^{2}=0,
$$

which simplifies to $(a b+a c)^{2}+4(a b+a c)+4=0$, or equivalently $[(a b+a c)+2]^{2}=0$.
Thus, $a b+a c=-2$.