



 
Hi Rick, Suppose the three integers are $a, b$ and $c,$ then since their product is negative either one of them is negative and the other two are positive or all three are negative. When dealing with the product of integers it is quite often helpful to consider the prime factors so I factored 24 to get \[a \times b \times c = 24 = 2^3 \times 3.\] Hence the only primes that divide $a, b$ and $c$ are 2 and 3. From this I can see that the divisors of 24 are 1 and 24, 2 and 12, 3 and 8, and 4 and 6. These are the only possible choices for $a, b$ and $c.$ If exactly one of $a, b$ and $c$ is negative then it must be 12 less than the sum of the other 2. I didn't see how to do that. If all three are negative then since their sum is 12 I looked through the possibilities and saw that 8, 3, and 1 will work. I hope this helps, 



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