Giải thích các bước giải:
Ta có : $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}$
$\to \dfrac{x-1}{2} = \dfrac{2y-4}{6} = \dfrac{3z-9}{12} = \dfrac{x-1-(2y-4)+(3x-9)}{2-6+12} = \dfrac{(x-2y+3z)-6}{8} = \dfrac{-10-6}{8} =-2$
$\to \left\{ \begin{array}{l}\dfrac{x-1}{2} = -2\\\dfrac{y-2}{3} = -2\\\dfrac{z-3}{4} = =2\end{array} \right.$
$\to x = -3,y=-4,z=-5$
Vậy $(x,y,z) = (-3,-4,-5)$