Đặt k: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)K
=> x = 2k + 1
=> y = 3k + 2
=> z = 4k + 3
Thay vào: x - 2y + 3z = 14
=> 2k + 1 - 2 * ( 3k + 2) + 3 * (4k + 3) = 14
=> 2k + 1 - 6k + 4 + 12k + 9 =14
=> ( 2k - 6k + 12k) + ( 1 + 4 + 9) = 14
=> 8k + 14 = 14
=> 8k = 14 -14 = 0
=> k = 0 / 8 = 0
\(=>\frac{x-1}{2}=0=>x=0\cdot2+1=1\)
\(=>\frac{y-2}{3}=0=>y=0\cdot3+2=2\)
\(=>\frac{z-3}{4}=0=>z=0\cdot4+3=3\)
Vậy x = 1
y = 2
z = 3