3. Given the parabola $y=a x^{2}+b x+c$ passes through two points $A(-1,2)$ and $B(-2,3)$, and does not pass through the first quadrant. If $s=a+b-c$, then the range of $s$ is ( ).
(A) $s \leqslant-3$
(B) $s<2$
(C) $s \leqslant 2$
(D) $s<-3$.
3.A.

From the problem, we have $\left\{\begin{array}{l}a-b+c=2, \\ 4 a-2 b+c=3 \text {. }\end{array}\right.$
Solving, we get $a=\frac{b}{3}+\frac{1}{3}, c=\frac{2}{3} b+\frac{5}{3}$.
Since the parabola does not pass through the first quadrant, we know $a<0, c \leqslant 0$, i.e., $\square$
$$
\frac{b}{3}+\frac{1}{3}<0, \frac{2}{3} b+\frac{5}{3} \leqslant 0 .
$$

Solving, we get $b \leqslant-\frac{5}{2}$.
Thus, $s=a+b-c$
$$
\begin{array}{l}
=\left(\frac{b}{3}+\frac{1}{3}\right)+b-\left(\frac{2}{3} b+\frac{5}{3}\right) \\
=\frac{2}{3} b-\frac{4}{3} \leqslant-3 .
\end{array}
$$