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?

Hint $ $ Recall $ \ d = \gcd(a,b,c)\,\Rightarrow\, \color{#c00}d\mid a,b,\color{#c00}c,\ $ i.e. the $\rm\,g\color{#0a0}{cd}$ is a $\rm\color{#0a0}{c}ommon\ \color{#0a0}{d}ivisor$ of its arguments.
In your case $ \ d = \gcd(a,b)\,$ thus $\, \color{#c00}{d}\mid \color{#c00}c\, $ becomes $\,\gcd(a,b)\mid c,\,$ as desired
