Trigonometry and Limits:

First let's solve the limit: by L'Hopital , $\begin{aligned} \lim_{x\to a} \frac{\sqrt{x^2-a^2}}{\sqrt{x-a} + \sqrt{x} - \sqrt{a}} &= \lim_{x\to a} \frac{\frac{x}{\sqrt{x^2-a^2}}}{\frac1{2\sqrt{x-a}} + \frac1{2\sqrt{x}}} \\ &= \lim_{x\to a} \frac{x}{\frac{\sqrt{x+a}}2 + \frac{\sqrt{x^2-a^2}}{2\sqrt{x}}} \\ &= \frac{a}{\frac{\sqrt{2a}}2 + 0} = \sqrt{2a}. \end{aligned}$

Now let $z_1 = e^{i\alpha}, z_2 = e^{i\beta}, z_3 = e^{i\gamma}.$ Then the two equations taken together give $\begin{aligned} z_1 + z_2 + z_3 &= 2z_1z_2z_3 \\ {\overline{z_1}} + {\overline{z_2}} + {\overline{z_3}} &= 2{\overline{z_1z_2z_3}} \end{aligned}$ and multiplying the second equation by $z_1z_2z_3$ gives $z_2z_3 + z_1z_3 + z_1z_2 = 2$ but the left side of this equation is just $\cos(\alpha+\beta) + \cos(\gamma+\alpha) + \cos(\beta+\gamma),$ so $a=2,$ and the answer is $\sqrt{2a} = \fbox{2}.$