\(\dfrac{2bz-3cy}{a}=\dfrac{3cx-az}{2b}=\dfrac{ay-2bx}{3c}\\ \Rightarrow\dfrac{2abz-3acy}{a}=\dfrac{6bcx-2abz}{2b}=\dfrac{3acy-6bcx}{3c}\\ =\dfrac{\left(2abz-3acy\right)+\left(6bcx-2abz\right)+\left(3acy-6bcx\right)}{a+2b+3c}\\ =\dfrac{\left(2abz-2abz\right)+\left(3acy-3acy\right)+\left(6bcx-6bcx\right)}{a+2b+3c}=0\\ \)
\(\Rightarrow2bz-3cy=3cx-az=ay-2bx=0\\ \Rightarrow\left\{{}\begin{matrix}2bz=3cy\\3cx=az\\ay=2bx\end{matrix}\right.\)
\(2bz=3cy\Rightarrow\dfrac{2b}{y}=\dfrac{3c}{z}\\ 3cx=az\Rightarrow\dfrac{3c}{z}=\dfrac{a}{x}\\ ay=2bx\Rightarrow\dfrac{a}{x}=\dfrac{2b}{y}\\ \Rightarrow\dfrac{a}{x}=\dfrac{2b}{y}=\dfrac{3c}{z}\Rightarrow.....\)
