Ta có: `(2bz-3cy)/a` = `(3cx-az)/(2b)` = `(ay-2bx)/(3c)`
`=>` $\frac{4bzx-6cyx}{2ax}$ = $\frac{6cxy−2azy}{4by}$ = $\frac{2ayz−4bxz}{6cz}$ = $\frac{4bzx−6cyx+6cxy− 2azy+2ayz−4bxz}{2ax+4by+6zy}$
`=>` $\frac{0}{2ax+4by+6zy}$ = 0
`=>` $2bz$ $=$ $3cy$ $;$ $3cx$ $=$ $az$ $;$ $ay$ $=$ $2bx$
`=>` `y/(2b)` = `z/(3c)` ; `x/a` = `z/(3x)` ; `x/a` = `y/(2b)`
`=>` `x/a` = `y/(2b)` = `z/(3c)` ( đpcm )
