How can I prove that $10=2^{a}*3^{b}*7^{c}$ has infinite solutions?

Question

Both in a unrescrited case and with the following restriction: $a+b+c=1$

Answer 1

For the unrestricted case, let $a=1.$ Now you need $3^b7^c=5$, so for a positive integer $n$, let

$$3^b = 5^{n+1}, 7^c = 5^{-n}$$.

Solve these two equations for $b$ and $c$.

That should give you a leg up on the restricted case.

Answer 2

Hint $$c=1-a-b$$

Your equation then becomes $$\frac{10}{7}=(\frac{2}{7})^a (\frac{3}{7})^b$$

Show now that if you fix $a$ you can find some $b$ which satisfies this equation.