32at8 ** Let $a, b, c \in \mathbf{R}^{+}, a+b+c=1$, find the minimum value of $\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-$ $\frac{8 c}{a+b+3 c}$.
Let $\left\{\begin{array}{l}a+2 b+c=x, \\ a+b+2 c=y, \\ a+b+3 c=z,\end{array}\right.$ then $\left\{\begin{array}{l}x-y=b-c, \\ z-y=c,\end{array}\right.$ so $\left\{\begin{array}{l}a+3 c=2 y-x, \\ b=z+x-2 y, \\ c=z-y,\end{array}\right.$
$$
\begin{aligned}
\text { Hence the original expression } & =\frac{2 y-x}{x}+\frac{4(z+x-2 y)}{y}-\frac{8(z-y)}{z} \\
& =-17+2 \frac{y}{x}+4 \frac{x}{y}+4 \frac{z}{y}+8 \frac{y}{z} \\
& \geqslant-17+2 \sqrt{8}+2 \sqrt{32}=-17+12 \sqrt{2} .
\end{aligned}
$$

When $a=3-2 \sqrt{2}, b=\sqrt{2}-1, c=\sqrt{2}$, the equality holds. Therefore, the required minimum value is $-17+$ $12 \sqrt{2}$.