If positive integer n is divisible by both 4 and 21, then n must be divisible by which of the following?

the answer is 12.

the prime factors of 4 is 2.
the prime factors of 21 is 3, 7

the prime factors of 12 is 2,3,2

so what happens to the 7??? why is the 7 left out but the 3 is kept?

you should have included the answer choices, for a fuller discussion of the problem.

if n is divisible by some number, all that means is that the number in question goes into n. there is no requirement that the number contain ALL of n's prime factors!

so, in this case, n is divisible by 4 (= 2 x 2) and 21 (= 3 x 7). since there is no overlap between these prime factorizations, it follows that n is divisible by 2 x 2 x 3 x 7.
this means that n MUST be divisible by ANY AND ALL of the following 12 numbers:
1
2
3
7
4 (= 2 x 2)
6 (= 2 x 3)
14 (= 2 x 7)
21 (= 3 x 7)
12 (= 2 x 2 x 3)
28 (= 2 x 2 x 7)
42 (= 2 x 3 x 7)
84 (= 2 x 2 x 3 x 7)
in other words, all the possible subsets of two 2's, one 3, and one 7.
the fact that the correct answer is 12 means only that 12 is the only one of these twelve possibilities that actually showed up in the answer choices; it doesn't indicate that 12 is special in any way, or that we're forsaking the "7" for any particular reason. the correct answer could just as well have been any one of the other eleven numbers in this list.