$$\text{Re}\left(\frac{z_1 + z_2}{z_3}\right) = -1$$
Let $$z_1 + z_2 = z_s$$ (the resultant vector of $$z_1$$ and $$z_2$$). The equation becomes:
$$\text{Re}\left(\frac{z_s}{z_3}\right) = -1 \implies \frac{|z_s|}{|z_3|} \cos(\theta) = -1$$
Where $\theta$ is the angle between the vector $$(z_1 + z_2)$$ and $$z_3$$. Since $$|z_3| = 2$$:
$$|z_1 + z_2| \cos(\theta) = -2$$
For any triangle inscribed in a circle, the product of its sides $$|z_1 - z_2| \cdot |z_2 - z_3| \cdot |z_3 - z_1|$$ is maximized when the triangle is equilateral.
In an equilateral triangle, the angle between any two position vectors from the origin is $$120^\circ$$.
The resultant $$|z_1 + z_2|$$ of two vectors of length 2 at $$120^\circ$$ is exactly $$2$$
The vector $$(z_1 + z_2)$$ points directly opposite to the third vertex $$z_3$$, meaning the angle $$\theta$$ between them is $$180^\circ$$.
$$|z_1 + z_2| \cos(\theta) = 2 \cdot \cos(180^\circ) = 2(-1) = -2$$
For an equilateral triangle inscribed in a circle of radius R = 2, the side length s is:
$$s = R\sqrt{3} = 2\sqrt{3}$$The required product is simply $$s^3$$
$$\text{Maximum Value} = (2\sqrt{3})^3 = 8 \cdot 3\sqrt{3} = 24\sqrt{3}$$
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