Theo bài ra ta có : \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
\(\Rightarrow\frac{a\left(2bz-3cy\right)}{a^2}=\frac{2b\left(3cx-az\right)}{\left(2b\right)^2}=\frac{3c\left(ay-2bx\right)}{\left(3c\right)^2}\)
\(\Rightarrow\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{\left(2b\right)^2}=\frac{3acy-6bcx}{\left(3c\right)^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{\left(2b\right)^2}=\frac{3acy-6bcx}{\left(3c\right)^2}=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+\left(2b\right)^2+\left(3c\right)^2}=0\)
=> \(\hept{\begin{cases}2bz=3cy\\3cx=az\\ay=2bx\end{cases}}\Rightarrow\hept{\begin{cases}\frac{z}{3c}=\frac{y}{2b}\\\frac{z}{3c}=\frac{x}{a}\\\frac{y}{2b}=\frac{x}{a}\end{cases}\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\left(\text{đpcm}\right)}\)