Real Number Problem: Evaluate xyz with Given Equations

[anemone]

If $x,\,y,\,z$ are real numbers such that $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, evaluate $xyz$.

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[anemone]

Congratulations to Opalg for his correct solution, which is shown below:):

Spoiler

The equation with roots $x$, $2y$, $3z$ must be of the form $\lambda^3 - 6\lambda^2 + 12\lambda + c = 0$ for some constant $c$, or $(\lambda-2)^3 + c + 8 = 0.$ But the function $(\lambda-2)^3$ is strictly increasing everywhere except at $\lambda = 2.$ So the only way that the equation $(\lambda-2)^3 + c + 8 = 0$ can have three real roots is if $c = -8$ (and then the three roots are all equal to $2$). Therefore $x=2$, $y=1$, $z=\frac23$ and their product is $xyz = \frac43.$