Three, 11. A. Let the equation $x^{2}+a x+b=0$ have two roots $\alpha, \beta (\alpha, \beta$ are integers, and $\alpha \leqslant \beta)$. Then the roots of the equation $x^{2}+\alpha x+a=0$ are $\alpha+1, \beta+1$. From the problem, we have $\alpha+\beta=-a, (\alpha+1)(\beta+1)=a$.
Adding the two equations gives $\alpha \beta+2 \alpha+2 \beta+1=0$, which simplifies to $(\alpha+2)(\beta+2)=3$.
Therefore, $(\alpha+2, \beta+2)=(1,3)$ or $(-3,-1)$.
Solving gives $(\alpha, \beta)=(-1,1)$ or $(-5,-3)$.
Thus, $a+b+c$
$$\begin{array}{l} =-(\alpha+\beta)+\alpha \beta-[(\alpha+1)+(\beta+1)] \\ =-3 \text { or } 29 . \end{array}$$