11. (Full marks 12 points) Given real numbers $a$ and $b$ satisfy $a^{2} + ab + b^{2} = 1$. Find the range of $a^{2} - ab + b^{2}$.
Three, let $a^{2}-a b+b^{2}=k$,
and $a^{2}+a b+b^{2}=1$.
From (1) and (2), we solve to get $a b=\frac{1}{2}(1-k)$.
Thus, we have $(a+b)^{2}=\frac{1}{2} \cdot(3-k)$.
$$
\because(a+\dot{b})^{2} \geqslant 0, \therefore k \leqslant 3 \text {. }
$$

Therefore, $a+b= \pm \sqrt{\frac{3-k}{2}}$.
Hence, $a$ and $b$ are the two real roots of the equation $t^{2} \mp \sqrt{\frac{3-k}{2}} t+\frac{1-k}{2}=0$,
$$
\therefore \Delta=\frac{1}{2}(3-k)-4 \times \frac{1}{2}(1-k) \geqslant 0 .
$$

Thus, $\frac{1}{3} \leqslant k \leqslant 3$. That is, $\frac{1}{3} \leqslant a^{2}-a b+b^{2} \leqslant 3$.