
3. Determine all triples $(x, y, z)$ consisting of three distinct real numbers, that satisfy the following system of equations:

$$
\begin{aligned}
x^{2}+y^{2} & =-x+3 y+z, \\
y^{2}+z^{2} & =x+3 y-z, \\
x^{2}+z^{2} & =2 x+2 y-z .
\end{aligned}
$$


3. If we subtract the second equation from the first, we obtain

$$
x^{2}-z^{2}=-x+z-(x-z),
$$

which can be rewritten as

$$
(x-z)(x+z)=-2(x-z),
$$

and then simplified further to get

$$
(x-z)(x+z+2)=0 \text {. }
$$

Since $x$ and $z$ must be different, $x-z$ is nonzero and we can conclude that $x+z+2=0$, hence $z=-x-2$. If we subtract the third equation from the second, we obtain

$$
y^{2}-x^{2}=x+3 y-(2 x+2 y),
$$

which can be rewritten as

$$
(y-x)(y+x)=1(y-x),
$$

and then simplified further to get

$$
(y-x)(y+x-1)=0 \text {. }
$$

Since $x$ and $y$ must be different, $y-x$ is nonzero and we can conclude that $y+x-1=0$, hence $y=1-x$.

If we now substitute $y=1-x$ and $z=-x-2$ in the first equation, we get

$$
x^{2}+(1-x)^{2}=-x+3(1-x)+(-x-2) \text {, }
$$

which we can rewrite as

$$
2 x^{2}-2 x+1=-5 x+1 \text {. }
$$

This is a quadratic equation $2 x^{2}+3 x=0$, whose roots are $x=0$ and $x=-\frac{3}{2}$. In both cases we can deduce the values of $y$ and $z$ from the formulas $y=1-x$ and $z=-x-2$. This gives the solutions $(x, y, z)=\left(-\frac{3}{2}, \frac{5}{2},-\frac{1}{2}\right)$ and $(x, y, z)=(0,1,-2)$.

For these solutions we know that they satisfy the first equation. Since the difference between the first and the second equation gives zero on both sides (since we chose $z=-x-2$ ), the second equation is satisfied as well. Because of the choice $y=1-x$, also the third equation is satisfied. We conclude that both solutions indeed satisfy all three equations.

