Here we have to do the decomposition carefully..
Let us have S(A,B,C) that contains the following FDs :
AB --> C
BC --> A
AC --> B
Hence all the determinants(lhs terms) are candidate keys and hence superkeys..And hence it follows BCNF rule.
Now we have one fd remaining D --> E so let us have that in T(D,E) ..having D as the candidate key and hence superkey..
But still we have the attribute F which is not yet covered .But we cannot include them in either S or T subrelation as that will violate the BCNF rule as F has to be a part of superkey and hence BCNF form will be violated if we try to preserve FDs of R relation..And if we try to preserve BCNF property , then the FDs will be violated..Hence we have to place the attribute F in separate table , say W..
Now to make lossless we add the attributes A,B or A,C or B,C and D to W to make it W(A,B,D,F) or W(A,C,D,F) or W(B,C,D,F) and this subrelation is also in BCNF as it has no non trivial FDs..
So W table will have 2 foreign keys one referencing to S and other referencing to T subrelation..
So no of foreign keys required = 2
Hence B) is the correct option..
