Từ \(\left\{\begin{matrix}2\left(x-3\right)=3\left(y+2\right)\\5\left(2-z\right)=3\left(y+2\right)\end{matrix}\right.\)\(\Rightarrow2\left(x-3\right)=3\left(y+2\right)=5\left(2-z\right)\)
\(\Rightarrow\frac{2\left(x-3\right)}{30}=\frac{3\left(y+2\right)}{30}=\frac{5\left(2-z\right)}{30}\)\(\Rightarrow\frac{x-3}{15}=\frac{y+2}{10}=\frac{2-z}{6}\)
Đặt \(\frac{x-3}{15}=\frac{y+2}{10}=\frac{2-z}{6}=k\) suy ra \(\left\{\begin{matrix}x=15k+3\\y=10k-2\\z=2-6k\end{matrix}\right.\)
\(\Rightarrow2x-3y+z=-4\Rightarrow2\left(15k+3\right)-3\left(10k-2\right)+2-6k=-4\)
\(\Rightarrow30k+6-30k+6+2-6k=-4\)
\(\Rightarrow14-6k=-4\Rightarrow6k=18\Rightarrow k=3\)
Suy ra \(\left\{\begin{matrix}x=15k+3=15\cdot3+3=48\\y=10k-2=10\cdot3-2=28\\z=2-6k=2-6\cdot3=-16\end{matrix}\right.\)
Giá trị của \(x-y+z=48-28+\left(-16\right)=4\)