oooo sorry incorrect - cannot split into 6 and 6 - one of the people is lighter *OR* heavier , splitting 6 and 6 cannot reveal whether the heavy/light person is on either side ... Wink ... (the heavy side of 6 could reveal that those 3 vs 3 of them are equal on the 2nd weighing , and leave you with an inability to determine who's the odd duck out of the 6 who weren't on the heavy side of the 1st weighing) ... it's a tricky one ... 6v6 was my first inclination as well , but the key is 1 of 12 is lighter OR heavier ...
In this case you will have to split the odds:
Heavier person case:
1. You split them in half (six persons per side)
2. Take the heaviest side and split them once more (3 persons per side)
3. Take the heaviest side and you end up with 3 persons.
4. Choose randomly the 2 of them and weight them.
If they're equal, you found the one outside. Otherwise the heavier one will be revealed by the balance.
Thinner person case:
Then you have to repeat the process for the case the person is thinner than the others.
1. You split them in half (six persons per side)
2. Take the lightest side and split them once more (3 persons per side)
3. Take the lightest side and you end up with 3 persons.
4. Choose randomly the 2 of them and weight them.
If they're equal, you found the one outside. Otherwise the thinner one will be revealed by the balance.
There is another way by dividing the weight of the persons you weighted and find if they're of the same weight every time, (ie: first time total weight/6, second time total weight/3 etc), but... I prefer simpler solutions
PS: It seems I missed the job then