Let the digits of $p$ be $a, b$, the digits of $q$ be $c, d$, the digits of $r$ be $e, f$. Taking the equations in pairs it's evident that $(10a+b)(c+d)=(a+b)(10c+d)$ (other cases are similar) so $ad=bc$, $af=be$, $cf=de$. We now do casework on whether the lowest digit is a first digit or a second digit. Case $1$. lowest digit is a first digit, WLOG let it be $a$ Sub-case $1.1$: $a=1$ $d=bc$, $f=be$. So $f$ and $d$ must be multiples of $b$. The only possibility that works is $b=2$, $c=3$, $e=4$, $d=6$, $f=8$. (Of course we can switch $c$ and $e$ but those are just permutations of the same three numbers.) So $(12, 36, 48)$ works. Sub-case $1.2$: $a=2$ $2d=bc$, $2f=be$. Suppose $b=3$. Then both $2d=3c$ and $2f=3e$ are true, and now we don't have enough numbers to satisfy this for distinct $c,d,e,f$. Same is true for $b=5$. Suppose $b=4$. Then both $d=2c$ and $f=2e$ are true. But again, since we already took $2$ and $4$, we don't have enough numbers to satisfy this. We can see for $a>2$ we'll need to use either $5$ or $7$ which can't work because there aren't any other corresponding digits with $5$ or $7$ as a factor. So case $1$ finished Case $2$: lowest digit is a second digit, WLOG let it be $b$ Sub-case $2.1$: $b=0$ We easily see this is impossible because $ad=0$, but we can't have two of the same digit. Rest is similar to Case $1$, so we can also have $a=2$, $d=3$, $f=4$, $c=6$, $f=8$, so $21, 63, 84$ works. We can easily verify both possibilities work: the first gives $p \cdot q \cdot s(r) = p\cdot s(q) \cdot r = s (p) \cdot q \cdot r = 5184$ and the second gives $p \cdot q \cdot s(r) = p\cdot s(q) \cdot r = s (p) \cdot q \cdot r = 15876$. So the possible $p, q, r$ are $(12, 36, 48), (21, 63, 84)$ and permutations.