8. Given positive integers $a, b, c, x, y, z$ satisfying $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$,
and $\left\{\begin{array}{l}2 a+b+4 c=4 x y z, \\ 2 x+y+4 z=4 a b c .\end{array}\right.$
Then the number of six-tuples $(a, b, c, x, y, z)$ that satisfy the conditions is $\qquad$ groups.
8. 0 .

Assume $x \geqslant a$. Then
$$
\begin{array}{l}
4 x y z=2 a+b+4 c \leqslant 7 a \leqslant 7 x \\
\Rightarrow y z \leqslant \frac{7}{4} \Rightarrow y z \leqslant 1 \Rightarrow(y, z)=(1,1) \\
\Rightarrow\left\{\begin{array}{l}
a a+b+4 c=4 x, \\
2 x+5=4 a b c
\end{array}\right. \\
\Rightarrow 2 a+b+4 c+10=8 a b c .
\end{array}
$$

If $b \geqslant 2$, then
$$
2 a+b+4 c+10 \leqslant 12 a<8 a b c,
$$

a contradiction.
If $b=1$, then $\left\{\begin{array}{l}2 a+5=4 x, \\ 2 x+5=4 a\end{array}\right.$ has no positive integer solutions.