The 66th question: Find the smallest positive real number $\lambda$, such that for any three complex numbers $\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3} \in\{\mathrm{z} \in \mathrm{C} \| \mathrm{z} \mid<1\}$, if $\mathrm{z}_{1}+$ $z_{2}+z_{3}=0$, then $\left|z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right|^{2}+\left|z_{1} z_{2} z_{3}\right|^{2}<\lambda_{。}$
The 66th question,
Solution: Let $\mathrm{a}=\left|\mathrm{z}_{1}\right|^{2}, \mathrm{~b}=\left|\mathrm{z}_{2}\right|^{2}, \mathrm{c}=\left|\mathrm{z}_{3}\right|^{2}$, then
$$
\begin{array}{l}
=\sum \mathrm{bc}+\sum \mathrm{a}\left(\mathrm{z}_{2} \cdot \overline{\mathrm{z}_{3}}+\mathrm{z}_{3} \cdot \overline{\mathrm{z}_{2}}\right) \\
=\sum b c+\sum a\left[\left(z_{2}+z_{3}\right)\left(\overline{z_{2}}+\overline{z_{3}}\right)-z_{2} \overline{z_{2}}-z_{3} \overline{z_{3}}\right] \\
=\sum b c+\sum a\left[\left(-z_{1}\right)\left(-\overline{z_{1}}\right)-z_{2} \overline{z_{2}}-z_{3} \overline{z_{3}}\right] \\
=\sum \mathrm{bc}+\sum \mathrm{a}(\mathrm{a}-\mathrm{b}-\mathrm{c}) \\
=\sum a^{2}-\sum b c \\
\text { Also, }\left|z_{1} z_{2} z_{3}\right|^{2}=a b c \text {, so } \\
\left|z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right|^{2}+\left|z_{1} z_{2} z_{3}\right|^{2} \\
=\sum a^{2}-\sum b c+a b c \\
\leq \sum a-\sum b c+a b c \\
=1-(1-a)(1-b)(1-c) \\
<1 \\
\end{array}
$$

On the other hand, let $z_{1}=r, z_{2}=-r, z_{3}=0$, where $r \in \mathbb{R},|r|<1$, then $\left|z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right|^{2}+$ $\left|z_{1} z_{2} z_{3}\right|^{2}=r^{4}$, as $r \rightarrow 1$, we have $r^{4} \rightarrow 1$, hence $\lambda \geq 1$.
In conclusion, the minimum value of $\lambda$ is 1.