It will be 5.

Say 5 is included in any of the 3 triplets. Then the product of that triplet will necessarily end with 5 or 0 (obviously a multiple of 5).

Now, think - it is given that the product of the 3 triplets are equal. If 5 is included in one of the triplets - then the product of the other 2 triplets also has to end with 0 or 5 (to be a multiple of 5). This is impossible with the *unique* numbers 1-9, when 5 is already taken.

I just stressed on "ending with 0 or 5" to make it simple, but the accurate statement is "there is no way to form a multiple of 5 with the remaining numbers"

The odd one out in such problems will mostly be the prime number.