11. (20 points) Let $a_{i}, b_{i} \in \mathbf{C}$, satisfying $\left|a_{i}\right|^{2}+\left|b_{i}\right|^{2}=1(i=1,2,3)$. Find $\max _{i \neq j}\left\{\left|a_{i} \overline{a_{j}}+b_{i} \overline{b_{j}}\right|\right\}$'s minimum value.
11. Let $a_{i}=a_{1 i}, b_{i}=a_{2 i}(i=1,2,3)$,
$$
C_{k l}=\left|a_{1 k} \overline{a_{1 l}}+a_{2 k} \overline{a_{2 l}}\right|, c=\max _{k \neq l} C_{k l} \text {. }
$$

Then $6 c^{2}+3 \geqslant \sum_{k \neq l} C_{k l}^{2}+\sum_{k=l} C_{k l}^{2}$
$$
\begin{array}{l}
=\sum_{\substack{k=1 \\
l=1}}^{3}\left(a_{1 k} \overline{a_{1 l}}+a_{2 k} \overline{a_{2 l}}\right)\left(\overline{a_{1 k}} a_{1 l}+\overline{a_{2 k}} a_{2 l}\right) \\
=\sum_{\substack{k=1 \\
l=1}}^{3} \sum_{\substack{i=1 \\
j=1}}^{2} a_{i k} \overline{a_{j k}} \cdot \overline{a_{i l}} a_{j l} \\
=\sum_{\substack{i=1 \\
j=1}}^{2} \sum_{k=1}^{3} a_{i k} \overline{a_{j k}} \cdot \overline{a_{i l}} a_{j l} \\
=\sum_{\substack{i=1 \\
j=1}}^{2}\left|\sum_{k=1}^{3} a_{i k} \overline{a_{j k}}\right|^{2} \\
\geqslant \sum_{i=1}^{2}\left(\sum_{k=1}^{3} a_{i k} \overline{a_{i k}}\right)^{2} \\
=\left(\sum_{k=1}^{3}\left|a_{k}\right|^{2}\right)^{2}+\left(\sum_{k=1}^{3}\left|b_{k}\right|^{2}\right)^{2} \\
\geqslant \frac{\left(\sum_{k=1}^{3}\left(\left|a_{k}\right|^{2}+\left|b_{k}\right|^{2}\right)\right)^{2}}{2}=\frac{9}{2} .
\end{array}
$$

Thus, $c^{2} \geqslant \frac{1}{4}(c>0) \Rightarrow c \geqslant \frac{1}{2}$.
Let $a_{1}=b_{1}=\frac{\sqrt{2}}{2}, a_{2}=\overline{a_{3}}=\frac{1+\sqrt{3} \mathrm{i}}{2 \sqrt{2}}$,
$$
b_{2}=\overline{b_{3}}=\frac{1-\sqrt{3} \mathrm{i}}{2 \sqrt{2}} .
$$

It is easy to verify that, $\mathrm{C}_{k l}=\frac{1}{2}(k \neq l)$.
Therefore, $\max _{i \neq j}\left\{\left|a_{i} \bar{a}_{j}+b_{i} \overline{b_{j}}\right|\right\}$ has a minimum value of $\frac{1}{2}$.