Let $\cos A + \cos B + \cos C = x$ and $\sin A + \sin B + \sin C = y$. The cosine difference identity and some factoring on the left side of the given equation yields \begin{align*} \cos A \cos B+ \sin A \sin B + \cos B \cos C+ \sin B \sin C + \cos A \cos C+ \sin A \sin C &= 2(\frac{\cos A \cos B+ \sin A \sin B}{2}) + 2(\frac{\cos A \cos C+ \sin A \sin C}{2}) + 2(\frac{\cos C \cos B+ \sin C \sin B}{2}),\\ &= \frac{\cos C(x - \cos C) + \cos B(x - \cos B) + \cos A(x - \cos A) + \sin C(y - \sin C) + \sin B (y - \sin B) + \sin A(y - \sin A)}{2},\\ &= \frac{x^2 + y^2 - 3}{2}. \end{align*} Equating this value to $-\frac{3}{2}$, we find that $x^2 + y^2 = 0$. Since $A, B, C$ are all real angles, their corresponding cosine and sine values are also necessarily real. Thus, for the sum of the two squares to be $0$, both $x^2$ and $y^2$ also must be $0$. Thus $x^2 = x =\boxed{ \cos A + \cos B + \cos C= 0}$ and we are done.