Let $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=2002$, then $\left.\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right| z_{2} \right\rvert\,$ is

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Let $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=2002$, then $\left.\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right| z_{2} \right\rvert\,$ is
7. $\frac{1}{2002^{2}}$.

Since $z_{i} z_{1}^{-}=\left|z_{i}\right|^{2}=2002^{2}, i=1,2,3$. Therefore,
$$
\begin{array}{l}
\left|\frac{\frac{1}{z_{i}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}}{z_{1}+z_{2}+z_{3}}\right|=\frac{1}{2002^{2}}\left|\frac{\bar{z}_{1}}{z_{1}+\bar{z}_{2}+\bar{z}_{3}+z_{3}}\right| \\
=\frac{1}{2002^{2}} \cdot \frac{\left|\overline{z_{1}+z_{2}+z_{3}}\right|}{\left|z_{1}+z_{2}+z_{3}\right|}=\frac{1}{2002^{2}} . \\
\end{array}
$$