Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\frac{1}{x+y+z}=\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+z+x+2+x+y-3}{x+y+z}=\frac{2\left( x+y+z \right)}{x+y+z}=\frac{1}{2}$
Khi đó :
$\left\{ \begin{align} & y+z+1=2x \\ & z+x+2=2y \\ & x+y-3=2z \\ & x+y+z=2 \\ \end{align} \right.$$\Rightarrow \left\{ \begin{align} & x+y+z+1=3x \\ & x+y+z+2=3y \\ & x+y+z-3=3z \\ & x+y+z=2 \\ \end{align} \right.$$\Rightarrow \left\{ \begin{align} & 2+1=3x \\ & 2+2=3y \\ & 2-3=3z \\ \end{align} \right.$$\Rightarrow \left\{ \begin{align} & x=1 \\ & y=\frac{4}{3} \\ & z=\frac{-1}{3} \\ \end{align} \right.$
Vậy $\left( x;y;z \right)=\left( 1;\frac{4}{3};\frac{-1}{3} \right)$