Show that if there are integral solutions to the equation $ax+by+cz=e$, then $\gcd(a,b,c)\mid e$.
Now, suppose that $gcd(a,b,c)\mid e$. Show that there are integers $w$ and $z$ such that $\gcd(a,b)w + cz=e$.
Then show that there are integers $x$ and $y$ such that $ax+by=\gcd(a,b)w$.
For the first part, assume $gcd(a,b,c) = \alpha$. Then, by the division algorithm, $a = \alpha q$, $b = \alpha p$, $c = \alpha r$, for some integers $p,q,r$. The equation then becomes: $$\alpha px + \alpha qy + \alpha rz = e$$ dividing by $\alpha$, we see that $$px + qy + rz = \frac{e}{\alpha} $$ for the solutions to be integral, both sides of the equality must be integer valued, so $\alpha$ must divide $e$. As there wasn't really a question asked, I don't know which part you're struggling with. See if you can use this work to figure out the rest. Division algorithm will be your friend.