We have,
$$
\begin{aligned}
& \frac{27 x^2+32 x+16}{(3 x+2)^2(1-x)}=\frac{A}{3 x+2}+\frac{B}{(3 x+2)^2}+\frac{C}{1-x} \\
& \Rightarrow 27 x^2+32 x+16=A(3 x+2)(1-x)+B \\
& (1-x)+C(3 x+2)^2
\end{aligned}
$$
Put $x=1$ both sides, we get
$$
\begin{aligned}
27+32+16 & =0+0+25 C \\
C & =3
\end{aligned}
$$
Put $x=-2 / 3$ both sides, we get
$$
\begin{aligned}
12-\frac{64}{3}+16 & =\frac{5}{3} B \\
B & =4
\end{aligned}
$$
Put $x=0$ both sides, we get
$$
\begin{aligned}
16 & =2 A+B+4 C \\
2 A & =16-B-4 C=16-4-12=0 \\
& A=0 \\
\therefore \quad & A B+B C+C A=0+12+0=12
\end{aligned}
$$