a) Ta có : $\frac{x}{y}=\frac{3}{2}\Rightarrow \frac{x}{3}=\frac{y}{2}\Rightarrow \frac{x}{21}=\frac{y}{14}$
$5x=7z\Rightarrow \frac{x}{7}=\frac{z}{5}\Rightarrow \frac{x}{21}=\frac{z}{15}$
$\Rightarrow \frac{x}{21}=\frac{y}{14}=\frac{z}{15}$
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\frac{x}{21}=\frac{y}{14}=\frac{z}{15}=\frac{x-2y+z}{21-2.14+15}=\frac{32}{8}=4$
$\Rightarrow \left\{ \begin{align}
& x=4.21=84 \\
& y=4.14=56 \\
& z=4.15=60 \\
\end{align} \right.$Vậy $\left( x;y;z \right)=\left( 84;56;60 \right)$
b) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left( x+y+z \right)}{x+y+z}=2$
$\Rightarrow \left\{ \begin{align}
& y+z+1=2x \\
& x+z+2=2y \\
& x+y-3=2z \\
& x+y+z=\frac{1}{2} \\
\end{align} \right.$$\Rightarrow \left\{ \begin{align}
& x+y+z+1=3x \\
& x+y+z+2=3y \\
& x+y+z-3=3z \\
& x+y+z=\frac{1}{2} \\
\end{align} \right.$$\Rightarrow \left\{ \begin{align}
& \frac{1}{2}+1=3x \\
& \frac{1}{2}+2=3y \\
& \frac{1}{2}-3=3z \\
\end{align} \right.$$\Rightarrow \left\{ \begin{align}
& x=\frac{1}{2} \\
& y=\frac{5}{6} \\
& z=\frac{-5}{6} \\
\end{align} \right.$
Vậy $\left( x;y;z \right)=\left( \frac{1}{2};\frac{5}{6};\frac{-5}{6} \right)$