Since, $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|.$

$\Rightarrow \left|z_1-0\right|=\left|z_2-0\right|=\left|z_3-0\right|.$

Thus, the given points are equidistant from origin, and we know that the point which is equidistant from the vertices of a triangle is the circumcentre of the triangle.

$\therefore 0$ is the circumcentre of an equilateral $∆ABC$ and we also know that for an equilateral triangle the centroid and circumcentre are same.

$\therefore \frac{x_1+x_2+x_3}{3}=0=\frac{y_1+y_2+y_3}{3}$

where, $z_1=x_1+i{y}_1$.

$\Rightarrow \frac{x_1+x_2+x_3}{3}+i\left(\frac{y_1+y_2+y_3}{3}\right)=0$

$\Rightarrow \left(x_1+i{y}_1\right)+\left(x_2+i{y}_2\right)+\left(x_3+i{y}_3\right)=0$

$\Rightarrow z_1+z_2+z_3=0.$