Let p, q, r and s be distinct positive integers. Let p, q be odd and r, s be even. Consider the following statements:

1. ${(p–r)}^2\left(\mathrm{qs}\right)$ is even.

2. $(q–s)q^{2}s$ is even.

3. ${(q+r)}^2(p+s)$ is odd.

Which of the statements given above are correct?

The correct option is 1, 2 and 3.

Let us analyze the given conditions and verify each statement step-by-step.

We are given that $p$ and $q$ are odd positive integers, and $r$ and $s$ are even positive integers. Let us recall the basic properties of odd and even numbers:
- The sum or difference of two odd numbers is even.
- The sum or difference of two even numbers is even.
- The sum or difference of an odd number and an even number is odd.
- The product of any integer with an even number is always even.
- The product of two odd numbers is odd.
- Any positive integer power of an odd number is odd, and any positive integer power of an even number is even.

Statement 1: ${(p-r)}^2\left(qs\right)$ is even.
Let us examine the terms in this expression:
- $p$ is odd and $r$ is even. Therefore, their difference $(p-r)$ is odd.
- The square of an odd number, ${(p-r)}^2$, is also odd.
- $q$ is odd and $s$ is even. The product of an odd and an even number, $qs$, is even because $s$ is even.
- Now, multiplying the odd term ${(p-r)}^2$ by the even term $\left(qs\right)$ results in an even number.
Thus, Statement 1 is correct.

Statement 2: $(q-s)q^{2}s$ is even.
Let us examine the terms:
- $q$ is odd and $s$ is even, so their difference $(q-s)$ is odd.
- $q^2$ is the square of an odd number, so it is odd.
- $s$ is even.
- The expression is a product of these terms: $(q-s)\times q^2\times s$. Since $s$ is even, the product of any integers with $s$ must be even.
Thus, Statement 2 is correct.

Statement 3: ${(q+r)}^2(p+s)$ is odd.
Let us examine the terms:
- $q$ is odd and $r$ is even. Their sum $(q+r)$ is odd.
- The square of this sum, ${(q+r)}^2$, is also odd.
- $p$ is odd and $s$ is even. Their sum $(p+s)$ is odd.
- The product of two odd numbers, ${(q+r)}^2\times (p+s)$, is odd.
Thus, Statement 3 is correct.

Since all three statements (1, 2, and 3) are correct, the correct option is "1, 2 and 3".