1eck2 * Given the three vertices of $\triangle Z_{1} Z_{2} Z_{3}$ correspond to the complex numbers $z_{1}, z_{2}, z_{3}$, find its area, and use this to find the equation of the line passing through the two complex numbers $z_{1}$ and $z_{2}$.
Parse the area $S_{\Delta Z_{1} Z_{2} Z_{3}}$ of $\triangle Z_{1} Z_{2} Z_{3}$
$$
\begin{array}{l}
=\frac{1}{2} \cdot Z_{1} Z_{2} \cdot Z_{1} Z_{3} \cdot \sin \angle Z_{2} Z_{1} Z_{3} \\
=\frac{1}{2} \cdot\left|z_{2}-z_{1}\right| \cdot\left|z_{3}-z_{1}\right| \cdot \operatorname{Im} \frac{z_{3}-z_{1}}{z_{2}-z_{1}} \cdot\left|\frac{z_{2}-z_{1}}{z_{3}-z_{1}}\right| \\
=\frac{1}{2} \cdot \operatorname{Im}\left(\bar{z}_{1} z_{2}+\bar{z}_{2} z_{3}+\bar{z}_{3} z_{1}\right) \\
=\frac{\mathrm{i}}{4} \left\lvert\, \begin{array}{ccc}
1 & 1 & 1 \\
z_{1} & z_{2} & z_{3} \\
\overline{z_{1}} & \overline{z_{2}} & \overline{z_{3}}
\end{array} .\right.
\end{array}
$$
(1) and (2) in the expressions for the area given may be negative, because they represent directed areas, satisfying $S_{\triangle Z_{1} Z_{2} Z_{3}}=-S_{\triangle Z_{2} Z_{1} Z_{3}}$. If it is the usual area, take the modulus, which is $\left.\frac{1}{2} \right\rvert\, \operatorname{Im}\left(\overline{z_{1}} z_{2}\right.$ $\left.+\overline{z_{2}} z_{3}+\overline{z_{3}} z_{1}\right) \left.\left|=\frac{1}{4}\right|\left|\begin{array}{ccc}1 & 1 & 1 \\ z_{1} & z_{2} & z_{3} \\ \overline{z_{1}} & \overline{z_{2}} & \overline{z_{3}}\end{array}\right| \right\rvert\,$. The necessary and sufficient condition for three points $Z_{1}, Z_{2}, Z_{3}$ on the complex plane to be collinear is
$$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
z_{1} & z_{2} & z_{3} \\
\overline{z_{1}} & \overline{z_{2}} & \overline{z_{3}}
\end{array}\right|=0
$$

i.e., $z_{1} z_{2}+z_{2} \overline{z_{3}}+z_{3} \overline{z_{1}}=\overline{z_{1}} z_{2}+\overline{z_{2}} z_{3}+\overline{z_{3}} z_{1}$, thus the equation of the line $Z_{1} Z_{2}$ is: