It is known that:
If $\gcd(a,b)=1$, then $\gcd(ab,c)=\gcd(a,c) \cdot \gcd(b,c)$.
Let $p$ be a prime number such that $p\mid a$. Then $v_p(b)=0$ (since $\gcd(a,b)=1$), $\min\{v_p(b),v_p(c)\}=0$ and : $$\min\{v_p(ab),v_p(c)\}=\min\{v_p(a),v_p(c)\}=\min\{v_p(a),v_p(c)\}+\min\{v_p(b),v_p(c)\}$$ The same happens with $p\mid b$, thus $$v_p(\gcd(ab,c))=v_p(\gcd(a,c)) + v_p(\gcd(b,c))=v_p(\gcd(a,c) \cdot \gcd(b,c))$$ $$\implies \gcd(ab,c)=\gcd(a,c) \cdot \gcd(b,c)$$
However, if $\gcd(a,b)>1$, how can I calculate $\gcd(ab,c)$ ?