Ta có :
\(\frac{x+3}{5}=\frac{y-2}{3}=\frac{z-1}{7}=\frac{3x+9}{15}=\frac{5y-10}{15}=\frac{7z-7}{49}=\frac{3x+9+5y-10-7z+7}{15+15-49}\)
\(=\frac{3x+5y-7z+6}{-19}=\frac{32+6}{-19}=-2\)
=> x = ( - 2 ) . 5 - 3 = -13
y = ( - 2 ) . 3 + 2 = -4
z = ( - 2 ) . 7 + 1 = - 13
Vậy x = -13 ; y = -4 ; z = -13