Prove If 3 does not | n, then 3 | (n^2 + 3n -10)

I have to use proof by cases and cannot think of what to do. If 3 does not | n, then 3 | (n^2 + 3n -10) If someone could help me try and write this proof or at least do one case to guide me, I would appreciate it. Please explain what you did to or at least try. I have this one in my notes and I cannot figure out what the professor was doing. He mentions there are two cases.

Any integer n can be written as n = 3q + r, where r is 0, 1, or 2. Clearly, r is one or two in this case as 3 does not divide n.

Note that n^2 + 3n - 10 = (n + 5)(n - 2) = (3q + r + 5)(3q + r - 2). If r = 1, then 3 divides the first factor, and thus, the entire product. If r = 2, then 3 divides the second factor, and thus the product. This completes your proof.

Since 3 does not divide n, then n must be of the form

3k + 1 or 3k + 2. Consider each case:

3k + 1: n^2 + 3n - 10 = (3k + 1)^2 + 3(3k + 1) - 10

= 9k^2 + 15k - 6 = 3(3k^2 + 5k - 2). This is divisible by 3.

3k + 2: n^2 + 3n - 10 = (3k + 2)^2 + 3(3k + 2) - 10

= 9k^2 + 21k = 3(3k^2 + 7k), also divisible by 3

n^2 + 3n - 10 = (n - 2)(n + 5)

Since n does not divide 3, we indeed have 2 cases.

Either

n ≡ 1 ≡ -5 (mod 3)

or

n ≡ 2 (mod 3)

In the first case, 3|(n + 5) and, in the second case, 3|(n - 2). Therefore, 3 divides (n - 2)(n + 5) = n^2 + 3n - 10