Ta có \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(=\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)
\(=\frac{x-1-\left(2y-4\right)+3z-9}{2-6+12}\)
\(=\frac{x-1-2y+4+3z-9}{8}\)
\(=\frac{\left(x-2y+3z\right)-6}{8}=\frac{14-6}{8}=\frac{8}{8}=1\)
Có \(\frac{x-1}{2}=1\Rightarrow x-1=2\Rightarrow x=3\)
\(\frac{y-2}{3}=1\Rightarrow y-2=3\Rightarrow y=5\)
\(\frac{z-3}{4}=1\Rightarrow z-3=4\Rightarrow z=7\)