Subtracting the second equality from the first we notice that the left side is a square of a trinomial. Indeed, we get
$$(x - y + 2z)^2 = -y^2 + 6y - 9,$$
which can be rearranged into
$$(x - y + 2z)^2 + (y - 3)^2 = 0.$$
Since $x$, $y$ and $z$ are real numbers the expressions in the brackets must be 0. So, $y = 3$ and $x = y - 2z = 3 - 2z$. Plugging this into the first equation we get $2z^2 - 3z + 1 = 0$, or $(2z - 1)(z - 1) = 0$. We obtain the same by plugging $y = 3$ and $x = 3 - 2z$ into the second equation. So, $z = 1$ or $z = \frac{1}{2}$. The solutions are $(1, 3, 1)$ and $(2, 3, \frac{1}{2})$.