Step 1: Pull out the constant.
Write $(3+2x)^{-5}=3^{-5}\left(1+\dfrac{2x}{3}\right)^{-5}$. This puts it in the standard $(1+u)^{-n}$ shape.
Step 2: Recall the general term.
For $(1+u)^{-n}$ the general term is $(-1)^r\,{}^{n+r-1}C_r\,u^r$. Here $n=5$, so the binomial count is ${}^{r+4}C_r$.
Step 3: Write the term for our case.
So $T_{r+1}=3^{-5}(-1)^r\,{}^{r+4}C_r\left(\dfrac{2x}{3}\right)^r$.
Step 4: Pick the power we need.
We want $x^{12}$, so set $r=12$. The binomial count is ${}^{16}C_{12}$.
Step 5: Collect the constants.
The coefficient is $3^{-5}\,{}^{16}C_{12}\left(\dfrac{2}{3}\right)^{12}={}^{16}C_{12}\,\dfrac{2^{12}}{3^{5}\cdot3^{12}}$.
Step 6: Combine the powers of 3.
Since $3^5\cdot3^{12}=3^{17}$, the coefficient is ${}^{16}C_{12}\,\dfrac{2^{12}}{3^{17}}$. \[ \boxed{{}^{16}C_{12}\,\frac{2^{12}}{3^{17}}} \]