Example 11 Let $a b c \neq 0$, if $(a+2 b+3 c)^{2}=14\left(a^{2}+\right.$ $b^{2}+c^{2}$ ), then the value of $\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}$ is $\qquad$.
Considering the Cauchy-Schwarz inequality, we have
$$\left(1^{2}+2^{2}+3^{2}\right)\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+2 b+3 c)^{2}$$
The equality "=" holds when $a: b: c=1: 2: 3$, hence
$$\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=8 \text {. }$$