First, lets assume that a,b,c are all +ve to simplify the question.
The three numbers can have the following combinations:
118 --> with 3 rearrangements ( because 1 is repeated, the possibilities = \(\frac{3!}{2!} = 3\))
127 --> with 6 rearrangements (because all numbers are different, the possibilities = \(3! = 6\))
136 --> with 6 rearrangements
145 --> with 6 rearrangements
226 --> with 3 rearrangements
235 --> with 6 rearrangements
244 --> with 3 rearrangements
334 --> with 3 rearrangements

Total = 36
(the above part is similar to another interesting counting question mentioned here: (https://gmatclub.com/forum/in-how-many-ways-10-identical-chocolates-be-distributed-among-3-child-307384.html#p2375066)

However, each of a,b,c can be either +ve or -ve,
so the combined possibilities of the signs = 2*2*2 = 8
for details, (a,b,c) can be:
(+,+,+)
(+,+,-)
(+,-,+)
(-,+,+)
(+,-,-)
(-,+,-)
(-,-,+)
(-,-,-)
8 possibilities

so the total number of possibilities = 36*8 = 288
D
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Hello, could you please explain why you're doing 9C2 instead of 9C3?