$\left ( m - \frac 1n \right) \left( n - \frac 1p \right) \left( p - \frac 1m \right) = \frac{(mn-1)(np-1)(pm-1)}{mnp}$ $\implies$ $p \lvert mn-1 \implies mn \equiv 1 \pmod p$ $m \lvert np-1 \implies np \equiv 1 \pmod m$ $n \lvert pm-1 \implies pm \equiv 1 \pmod n$ Let $p = \frac{mn-1}{k}$ $k<m$ to preserve the ordering of $m,n,p$ $n(mn-1) \equiv k \pmod m$ $m(mn-1) \equiv k \pmod n$ $n+k \equiv 0 \pmod m$ $m+k \equiv 0 \pmod n$ $m+k < 2n \implies m+k=n$ Therefore $n < 2m$ Therefore $n+k < 3k \implies n+k = 2m \implies m+k+k=2m \implies m=2k$ $m\frac{mn-1}{k} \equiv 1 \pmod n \implies 2(mn-1) \equiv 1 \pmod n \implies -2 \equiv 1 \pmod n \implies n=3$ or $n=1$ $m \le n \implies m=2$ or $m=1$ This leaves us with three ordered pairs $(m,n)$ $(1,1), (1,3), (2,3)$ $(m,n)=(1,1) \implies A = 0$ $(m,n)=(1,3) \implies p = 1$, contradicting the ordering. $(m,n)=(2,3) \implies p = 5$, and $A=21$ Therefore, the two possible values of $A$ are $0$ and $21$

The product equals $ mnp-m-n-p+\frac{mn+np+pm-1}{mnp}$, or $ mnp\leq mn+np+pm-1$. If $ p\geq3$, it must hold $ 3m+3n\leq 2mn\leq 3m+3n-1$, absurd, yielding $ p=1$ or $ p=2$. If $ p=2$, then $ 2mn$ must divide $ mn+2m+2n-1$, and if their ratio is at least $ 2$, then $ 2mn+2m+2n\leq4mn\leq mn+2m+2n-1$ absurd, hence $ mn=2m+2n-1$, and $ m,n$ must clearly be odd. If $ n\geq5$, then $ 2m+2n+m\leq mn=2m+2n-1$, absurd, and $ n=3$, yielding $ m=5$. We then find the product equal to $ \frac{14}{3}\frac{5}{2}\frac{9}{5}=21$. If $ p=1$, then $ mn$ must divide $ mn+m+n-1$, or $ mn$ must divide $ m+n-1$. If $ n\geq2$, then $ m+n\leq mn\leq m+n-1$, absurd, yielding $ n=1$, or one of the factors is $ 0$ and the product is $ 0$. The only values are $ 0$ and $ 21$.