11. B. Given that for any real number $x$, the quadratic function $y=a x^{2}+b x+c$ satisfies
$$
x^{2}+2 x+2 \leqslant y \leqslant 2 x^{2}+4 x+3 \text {, }
$$

and when $x=9$, $y=121$. Find the value of $a+b+c$.
11. B. From the given conditions, we have
$$
(x+1)^{2}+1 \leqslant y \leqslant 2(x+1)^{2}+1 \text {. }
$$

Then \( y=a(x+1)^{2}+1 \) where \( 1 \leqslant a \leqslant 2 \).
When \( x=9 \), \( y=121 \), we get
$$
100 a+1=121 \Rightarrow a=\frac{6}{5} \text {. }
$$

Thus, \( y=\frac{6}{5}(x+1)^{2}+1=\frac{6}{5} x^{2}+\frac{12}{5} x+\frac{11}{5} \).
Therefore, \( a+b+c=\frac{29}{5} \).