write 6 integers which have a product .....?

You ask this question as if we have the answer and are just hiding it from you. This is not true. We would have to do all the work you don't want to do. But here's how you can do it for yourself:

Find the factors of 42. Don't just find the prime factors, but find all the factors.

Now try combinations of factors, including the repeating factors, until you get the 3 different sums the problem asks for.

See how easy that was? Fun, too.

n(n+a million) = 10506 n² + n - 10506 = 0 (n+103)(n-102) = 0 n = -103, 102 The integers are -102 and -103 The useful integers 102 and 103 additionally fulfill the circumstances of consecutive and product = 10506.