How come the same reaction can be balanced in two distinct ways?
As per Nicolau's suggestion, : 

$$\{(1-2)\times\color\red{4}\}-2$$

gives 

$$2\ce{XeO3}\to2\ce{Xe}+\ce{3O2}$$

Thus, I conjecture that the second and third reaction differ by the fact of decomposition of $\ce{XeO3}$

$2$ can be converted into $1$ by multiplying by $\color \red{\frac 3 2}$ and decomposing one of the $\ce{XeO3} $ molecules.

Wikipedia also says that $\ce{XeO3}$ is unstable.