To find the numbers $p, q, r$ that satisfy the given conditions, we follow these steps:

Start by examining the conditions provided in the problem:
1. $p \cdot q \cdot s(r) = p \cdot s(q) \cdot r$
2. $p \cdot s(q) \cdot r = s(p) \cdot q \cdot r$

Since each of $p, q, r$ is a two-digit number formed using six different digits, it follows that:
- $p = 10a + b$
- $q = 10c + d$
- $r = 10e + f$

The digit sum for each is:
- $s(p) = a + b$
- $s(q) = c + d$
- $s(r) = e + f$

### Setting Up Equations

Given the equality:
$$p \cdot q \cdot s(r) = p \cdot s(q) \cdot r$$
we can simplify to:
$$q \cdot s(r) = s(q) \cdot r$$

Similarly, using the second equality:
$$p \cdot s(q) \cdot r = s(p) \cdot q \cdot r$$
we can simplify to:
$$p \cdot s(q) = s(p) \cdot q$$

### Solving the System of Equations

From these simplified forms, we have:
$$q \cdot (e + f) = (c + d) \cdot r$$
$$p \cdot (c + d) = (a + b) \cdot q$$

### Check Possible Combinations

Given $p, q, r$ must utilize distinct digits, we will check possible combinations of such digits while solving:
- Assume $p = 12$, $q = 36$, and $r = 48$.

Check the digit sums:
- $s(p) = 1 + 2 = 3$
- $s(q) = 3 + 6 = 9$
- $s(r) = 4 + 8 = 12$

Substitute into the equality conditions:
$$12 \cdot 36 \cdot 12 = 12 \cdot 9 \cdot 48 \quad \text{and} \quad 12 \cdot 9 \cdot 48 = 3 \cdot 36 \cdot 48$$
Both satisfy the conditions.

Similarly, examine $p = 21$, $q = 63$, and $r = 84$.

Check the digit sums:
- $s(p) = 2 + 1 = 3$
- $s(q) = 6 + 3 = 9$
- $s(r) = 8 + 4 = 12$

Substitute into the equality conditions:
$$21 \cdot 63 \cdot 12 = 21 \cdot 9 \cdot 84 \quad \text{and} \quad 21 \cdot 9 \cdot 84 = 3 \cdot 63 \cdot 84$$
Both satisfy the conditions.

### Conclusion

The solutions are:
$$\boxed{(12, 36, 48), (21, 63, 84)}$$