3. Given that $a$, $b$, and $c$ are all integers, and satisfy
$$
a^{2}+b^{2}+c^{2}+3 < a b + 3 b + 2 c \text{.}
$$

Then the quadratic equation with roots $a+b$ and $c-b$ is ( ).
(A) $x^{2}-3 x+2=0$
(B) $x^{2}+2 x-8=0$
(C) $x^{2}-4 x-5=0$
(D) $x^{2}-2 x-3=0$
3. (D).

Since $a, b, c$ are integers and satisfy
$$
a^{2}+b^{2}+c^{2}+3 < a b + 3 b + 2 c \text{, }
$$

we have $a^{2}+b^{2}+c^{2}+3 \leqslant a b + 3 b + 2 c - 1$.
Rearranging and completing the square, we get
$$
\left(a-\frac{b}{2}\right)^{2} + 3\left(\frac{b}{2}-1\right)^{2} + (c-1)^{2} \leqslant 0 \text{. }
$$

Therefore, $a-\frac{b}{2}=0, \frac{b}{2}-1=0, c-1=0$.
Solving these, we get $c=1, b=2, a=1$.
Thus, $a+b=3, c-b=-1$.
Hence, the required equation is $x^{2}-2 x-3=0$.