Notice that no matter what $a$ is, $(x,y,z) = (0,0,0)$ will always be a solution, so we want this to be our only solution. First, substitute $y = -(x+z)$ into the second equation and it becomes $-(x+z)^2 + azx = 0$ or $x^2 - xz(a - 2) + z^2 = 0$. Let $t = \frac{x}{z}$, where $z \neq 0$, then the equation above is equivalent to $t^2 - t(a-2) + 1 = 0$. Since we want $(0,0,0)$ to be our only solution, then this equation must have no solution, so $\Delta = (a-2)^2 - 4 < 0$ $\implies a(a-4) < 0$ $\implies 0 < a < 4$ So we've reduced the possible values for $a$ down to the interval $(0,4)$.