A question on multinomial theorem using binomial theorem $(3x^2+2x+c)^{12}=\sum A_r x^r$

and 

$\frac{A_{19}}{A_5}=\frac 1 {2^7}$

Find $c$. I really have no idea what to do with this. This was on a test. I have studied only binomial theorem. So, please use it only(combinatorics approach is always welcome).
By comparing coefficients
$$
A_{19} = \frac{12!}{9! \cdot 1! \cdot 2!} \cdot 3^9 \cdot 2 \cdot c^2 + \frac{12!}{8! \cdot 3! \cdot 1!} 3^8 \cdot 2^3 \cdot c + \frac{12!}{7! \cdot 5! \cdot 0!} 3^7 \cdot 2^5 = 3^7\left(\frac{12!}{9! \cdot 1! \cdot 2!}18c^2 + \frac{12!}{8! \cdot 3! \cdot 1!}24c + \frac{12!}{7! \cdot 5! \cdot 0!} 32\right) \;,
$$
and
$$
A_5 = \frac{12!}{7! \cdot 5! \cdot 0!} 2^5 \cdot c^7 + \frac{12!}{8! \cdot 3! \cdot 1!} 3 \cdot 2^3 \cdot c^8 + \frac{12!}{9! \cdot 1! \cdot 2!} 3^2 \cdot 2 \cdot c^9 = c^7\left(\frac{12!}{9! \cdot 1! \cdot 2!}18c^2 + \frac{12!}{8! \cdot 3! \cdot 1!}24c + \frac{12!}{7! \cdot 5! \cdot 0!} 32\right) \;.
$$

Thus, by dividing, we get 
$$
\frac{A_5}{A_{19}} = \left(\frac{c}{3}\right)^7 \;,
$$
which implies $c = 6$ by the given condition.
By comparing coefficients
$$
A_{19} = \frac{12!}{9! \cdot 1! \cdot 2!} \cdot 3^9 \cdot 2 \cdot c^2 + \frac{12!}{8! \cdot 3! \cdot 1!} 3^8 \cdot 2^3 \cdot c + \frac{12!}{7! \cdot 5! \cdot 0!} 3^7 \cdot 2^5 = 3^7\left(\frac{12!}{9! \cdot 1! \cdot 2!}18c^2 + \frac{12!}{8! \cdot 3! \cdot 1!}24c + \frac{12!}{7! \cdot 5! \cdot 0!} 32\right) \;,
$$
and
$$
A_5 = \frac{12!}{7! \cdot 5! \cdot 0!} 2^5 \cdot c^7 + \frac{12!}{8! \cdot 3! \cdot 1!} 3 \cdot 2^3 \cdot c^8 + \frac{12!}{9! \cdot 1! \cdot 2!} 3^2 \cdot 2 \cdot c^9 = c^7\left(\frac{12!}{9! \cdot 1! \cdot 2!}18c^2 + \frac{12!}{8! \cdot 3! \cdot 1!}24c + \frac{12!}{7! \cdot 5! \cdot 0!} 32\right) \;.
$$

Thus, by dividing, we get 
$$
\frac{A_5}{A_{19}} = \left(\frac{c}{3}\right)^7 \;,
$$
which implies $c = 6$ by the given condition.