Hint $\rm\ mod\ a\!:\, b\equiv 2c,\ 3c\equiv -2b \equiv -4c\,\Rightarrow\, 7c\equiv 0.\,$ So $\rm\, \rm\color{#C00}{if\ \ 7\nmid a\ \ then}\,$ $\rm\,c\equiv0\,\Rightarrow\,b\equiv 2c\equiv 0.$
$\rm\color{#C00}{It\ fails\ if\,\ 7\:|\:a.}$ If $\rm\:a = 7c,\,$ let $\rm\:b = 2c.\:$ So $\rm\:a\:|\:b\!-\!2c = 0,\ a=7c\:|\:2b\!+\!3c = 7c,\,$ but $\rm\:a=7c\nmid c.$
Remark $\ $ Geometrically the result says that, modulo $\rm\:a,\:$ the lines $\rm\: y = 2x,\ 2y = -3x\:$ intersect only at the origin if $\rm\:7\nmid a,\:$ but otherwise they intersect elsewhere too, e.g. mod $7$ the lines coincide since $\rm\:2y = - 3x = 4x,\,$ so $\rm\:y = 2x\:$ by scaling by $4 \equiv 1/2$. Thus, being identical, their intersection is the entire line $\rm\:y = 2x,\:$ i.e. the $7$ points $\rm\:(x,2x)\ mod\ 7.$
If you know linear algebra, note that the system $\rm\:2x-y = 0 = 3x+2y\:$ has determinant $7.\,$ So if $7$ is coprime to $\rm\,a,\,$ then $7$ is invertible mod $\rm\,a,\,$ so we can use Cramer's rule to infer $\rm\:(x,y) = (0,0).$