Eye color is normally determined by polygenic inheritance, which means that genes at multiple loci produce various pigments that allow for a continuous spectrum of eye color. However, since you have stated that only three distinct possibilities exist, let's assume polygenic inheritance does not occur.
At this point, assuming there is only a single gene coding for eye color, we must consider what the possible alleles (types of genes) are in play. Two major possibilities remain: multiple alleles or incomplete dominance.
Let's first assume that their are multiple alleles, which means that there is an allele for brown, green, and blue eyes (similar to blood types). In this case, assuming that blue was recessive to green and brown, both parents would have to be carriers for a child to be blue-eyed. However, if it were dominant, then only one parent would have to carry the allele (assuming no codominance).
If, on the other hand, the alleles exhibited incomplete dominance, than many other possibilities are brought to the table. If blue were an intermediate for two other alleles, then a heterozygote child would be required, which would need parents carrying both alleles. However, if it were the dominant or recessive phenotype, than both parents would need to carry one of the alleles.
I am not sure how in depth an answer you were looking for. Maybe what you were asking was in basis Mendelian dominance (which is complicated by three phenotypes). So let's assume only green and blue are possible. Then, if blue is recessive, both parents need to be carriers to have a blue child. However, if it is dominant, only one parent needs to carry the allele.