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?

First divide $a,b,c$ by $\gcd(a,b)$ so that we can assume that $a,b$ (and therefore $a,b,c$) are relatively prime. 
Apply now Euclid's algorithm to $a,b$. This will give you equations:
$a=bq_1+r_1$, $b=q_2r_1+r_2$, ..., $r_n=q_{n+1}r_{n-1}+r_{n+1}$, where $r_{n+1}=1$ is the $\gcd(a,b)$.
Substituting backwards these equations we get $ax+by=1$. Multiply the whole equation by $c$ to get $a(cx)+b(cy)=c$.
