The given question shows 3 characters which means that it is a trihybrid cross.
When the two individuals having the same genotype as A a B b C c are crossed with each other, they form 16 gametes during gametogenesis that is 8 from each parent. The gametes formed are ABC, ABc, AbC, Abc, aBC, abC, aBc, abc. These gametes are crossed individually with each other which results in the formation of 64 offspring.
AaBbCc x AaBbCc
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The genotypic ratio comes out to be: 1:2:2:2:4:8:4:4:2:2:4:1:2:4:2:1:2:1:4:2:2:1:2:1:2:1
Out of these offspring produced only one offspring possess the homozygous dominant genotype which is AABBCC.
Thus, the probability of producing the offspring with the genotype AABBCC is 1/64.
To know more about trihybrid cross:
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