Example 4 Given $a, b, c \in \mathbf{R}_{+}$. Find
$$
\frac{a}{b+3 c}+\frac{b}{8 c+4 a}+\frac{9 c}{3 a+2 b}
$$

the minimum value.
Explanation: Substitute the denominator. Let
$$
b+3 c=x, 8 c+4 a=y, 3 a+2 b=z .
$$

Then $a=-\frac{1}{3} x+\frac{1}{8} y+\frac{1}{6} z$,
$$
\begin{array}{l}
b=\frac{1}{2} x-\frac{3}{16} y+\frac{1}{4} z, \\
c=\frac{1}{6} x+\frac{1}{16} y-\frac{1}{12} z . \\
\text { Hence } \frac{a}{b+3 c}+\frac{b}{8 c+4 a}+\frac{9 c}{3 a+2 b} \\
=\frac{1}{8}\left(\frac{y}{x}+\frac{4 x}{y}\right)+\frac{1}{6}\left(\frac{z}{x}+\frac{9 x}{z}\right)+ \\
\frac{1}{16}\left(\frac{4 z}{y}+\frac{9 y}{z}\right)-\frac{61}{48} .
\end{array}
$$

By the AM-GM inequality,
$$
\begin{aligned}
& \text { (1) } \geqslant \frac{1}{8} \times 4+\frac{1}{6} \times 6+\frac{1}{16} \times 12-\frac{61}{48} \\
= & \frac{47}{48} .
\end{aligned}
$$

Equality holds if and only if $y=2 x, z=3 x$.

Therefore, when $a=10 c, b=21 c$, the minimum value sought is $\frac{47}{48}$.