11. Given that $a$ and $b$ are real numbers, and $a^{2}+a b+b^{2}=3$. If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$.
Three, 11. Let $a^{2}-a b+b^{2}=k$. From
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=3, \\
a^{2}-a b+b^{2}=k
\end{array} \Rightarrow a b=\frac{3-k}{2} .\right.
$$

Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$$
=3+\frac{3-k}{2}=\frac{9-k}{2} \text {. }
$$

Since $(a+b)^{2} \geqslant 0$, i.e., $\frac{9-k}{2} \geqslant 0$, hence $k \leqslant 9$.
Therefore, $a+b= \pm \sqrt{\frac{9-k}{2}}, a b=\frac{3-k}{2}$.
Thus, the real numbers $a$ and $b$ can be considered as the two roots of the quadratic equation $x^{2} \mp \sqrt{\frac{9-k}{2}} x+\frac{3-k}{2}=0$. Then, $\Delta=\left(\sqrt{\frac{9-k}{2}}\right)^{2}-4 \times \frac{3-k}{2}=\frac{3 k-3}{2} \geqslant 0$.
Solving this, we get $k \geqslant 1$.
Therefore, $1 \leqslant k \leqslant 9$.
Thus, the maximum value of $a^{2}-a b+b^{2}$ is $m=9$, and the minimum value is $n=1$.
Hence, $m+n=10$.