Example 14 Given $a, b, c \in \mathbf{R}^{+}$, find the minimum value of $\frac{a}{b+3 c}+$ $\frac{b}{8 c+4 a}+\frac{9 c}{3 a+2 b}$.
Solution: By substituting the denominator, let $b+3c=x$, $8c+4a=y$, $3a+2b=z$,

then $a=-\frac{1}{3}x+\frac{1}{8}y+\frac{1}{6}z$, $b=\frac{1}{2}x-\frac{3}{16}y+\frac{1}{4}z$, $c=\frac{1}{6}x+\frac{1}{16}y-\frac{1}{12}z$.

Thus, $\frac{a}{b+3c}+\frac{b}{8c+4a}+\frac{9c}{3a+2b}=\frac{1}{8}\left(\frac{y}{x}+\frac{4x}{y}\right)+\frac{1}{6}\left(\frac{z}{x}+\frac{9x}{z}\right)+\frac{1}{16}\left(\frac{4z}{y}+\frac{9y}{z}\right)-\frac{61}{48}$.

By the AM-GM inequality, the above expression $\geqslant \frac{1}{8} \times 4 + \frac{1}{6} \times 6 + \frac{1}{16} \times 12 - \frac{61}{48} = \frac{47}{48}$.

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

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