exercise of gene linkage in genetics
You did right for most of the exercise, but you got confused on one of the (apparently) easiest things.
Try to reason in terms of chromosomes, and not merely of genotypes:
F1 inherit their chromosomes from two homozygous parents, so you know that they are AbC and aBc.
You can visualize the cross with the recessive homozygote as follows:
Ab C     ab c

-- -  X  -- -

aB c     ab c

The second parent will always pass abc, so in order to have an offspring with genotype aabbcc you only need that parent 1 passes abc.
Given that the parentals are Ab and aB and that RF is 12% (0.12), regarding the first two loci, the frequencies of the gametes are
Ab = 0.44 
aB = 0.44 
AB = 0.06 
ab = 0.06

Including locus C, the frequency of the gamete abc is then simply 0.06 * 0.5 (because c is unlinked), and that corresponds to 3%.