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 $\left( x;y;z \right)=\left( 1;\frac{4}{3};\frac{-1}{3} \right)$