Obviously, $c \geqslant 2$. When $c \geqslant 5$, $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c} \leqslant \frac{1}{5} \Rightarrow \frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right)$ $\geqslant\left(\frac{4}{5}\right)^{3}=\frac{64}{125}$, which is a contradiction. Therefore, $c=2,3,4$.
When $c=2$, $a b=(a-1)(b-1) \Rightarrow a+b=1$ has no solution;
When $c=3$, $3 a b=4(a-1)(b-1) \Rightarrow a b-4(a+b)+4=0 \Rightarrow(a-4)(b-4)=12$ $\Rightarrow a-4=12,6,4 \Rightarrow(a, b, c)=(16,5,3),(10,6,3),(8,7,3)$, which gives 3 solutions;
When $c=4$, $4 a b=6(a-1)(b-1) \Rightarrow a b-3(a+b)+3=0 \Rightarrow(a-3)(b-3)=6$ $\Rightarrow a-3=6,3 \Rightarrow(a, b, c)=(9,4,4),(6,5,4)$, which gives 2 solutions. Therefore, the answer is $B$.
6. Satisfy $a \geqslant b \geqslant c$ and $a b c=2(a-1)(b-1)(c-1)$, the number of positive integer solutions $(a, b, c)$ is $\qquad$
A. 4
B. 5
C. 6
D. 7