$\begin{aligned} & \text { } \frac{1}{a+b}+\frac{1}{c+a}=\frac{3}{a+b+c} \\ & \Rightarrow \quad \frac{a+b+c}{a+b}+\frac{a+b+c}{c+a}=3 \\ & \Rightarrow \quad 1+\frac{c}{a+b}+1+\frac{b}{c+a}=3 \\ & \Rightarrow \quad c(c+a)+b(a+b)=(a+b)(c+a) \\ & \Rightarrow \quad c^2+a c+a b+b^2=a c+a^2+b c+a b \\ & \Rightarrow \quad b^2+c^2-a^2=b c \\ & \therefore \quad \cos A=\frac{b^2+c^2-a^2}{2 b c}=\frac{b c}{2 b c}=\frac{1}{2} \\ & \therefore \quad A=60^{\circ} \\ & \text { Hence, } \sin A=\sqrt{3} / 2 \quad\end{aligned}$