if $gcd(a,b)=gcd(a,b,c)$ then I need to prove that $ax+by=c$ has solution in $\mathbb Z$ if $gcd(a,b)=gcd(a,b,c)$ then I need to prove that $ax+by=c$ has solution in $\mathbb Z$ that is: $gcd(a,b)|c$
but how can I prove it with the given hypothesis?

let $g = gcd(a,b) = gcd(a,b,c)$, $c=d*g$
it exists $u,v$ such as $a*u+b*v=g$, $a*u*d+b*v*d=g*d=c$
