2. For which values of the real parameter $a$, does the system

$$
\left\{\begin{array}{l}
x+y+z=\sqrt{3 a-1} \\
x^{2}+y^{2}+z^{2}=a \\
x y=z^{2}
\end{array}\right.
$$

have distinct real solutions?
Solution. For the system to have a solution, it is required that $3a-1 \geq 0$ and $a \geq 0$, i.e., $a \geq \frac{1}{3}$. The second equation of the system is equivalent to the equation $(x+y)^{2}-2xy+z^{2}=a$. Then, by substituting $x+y$ and $xy$ from the first and third equations, respectively, we obtain the equation $(\sqrt{3a-1}-z)^{2}-2z^{2}+z^{2}=a$, which is equivalent to the equation $2\sqrt{3a-1}z=2a-1$. The last equation has a solution in the real numbers if $a \neq \frac{1}{3}$, so $a>\frac{1}{3}$. Then $z=\frac{2a-1}{2\sqrt{3a-1}}$. By substituting $z=\frac{2a-1}{2\sqrt{3a-1}}$ into the first and second equations of the system, we obtain the system

$$
\left\{\begin{array}{l}
x+y=\frac{4a-1}{2\sqrt{3a-1}} \\
xy=\left(\frac{2a-1}{2\sqrt{3a-1}}\right)^{2}
\end{array}\right.
$$

Then $x$ and $y$ are solutions to the quadratic equation

$$
t^{2}-\frac{4a-1}{2\sqrt{3a-1}}t+\left(\frac{2a-1}{2\sqrt{3a-1}}\right)^{2}=0
$$

Since the system must have distinct real solutions, it is required that

$$
\left(\frac{4a-1}{2\sqrt{3a-1}}\right)^{2}-4\left(\frac{2a-1}{2\sqrt{3a-1}}\right)^{2}>0
$$

The last inequality is equivalent to the inequality $\frac{8a-3}{4(3a-1)}>0$, and since the denominator is positive, it follows that $8a-3>0$. Therefore, $a>\frac{3}{8}$. Finally, for $a>\frac{3}{8}$, the system has distinct real solutions.