Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2, and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 is

r1, r2, r3 ∈ {1, 2, 3, 4, 5, 6}
The condition ωr1 + ωr2 + ωr3 = 0 implies that r1, r2, r3 are of the form 3k, 3k+1, 3k+2 for some integer k.
For a fair die, the possible values are {1, 2, 3, 4, 5, 6}.
The numbers that are of the form 3k are {3, 6}
The numbers that are of the form 3k+1 are {1, 4}
The numbers that are of the form 3k+2 are {2, 5}
The number of ways to choose r1, r2, r3 such that ωr1 + ωr2 + ωr3 = 0 is given by the number of ways to choose one number from {3,6}, one number from {1,4} and one number from {2,5}.
This is given by 2C1 × 2C1 × 2C1 = 8
The total number of possible outcomes when throwing the die three times is 6 × 6 × 6 = 216
The required probability is the ratio of the number of favorable outcomes to the total number of outcomes:
Probability = (2C1 × 2C1 × 2C1) / (6 × 6 × 6) = 8/216 = 1/27
The numbers r1, r2, r3 must be such that they are of the form 3k, 3k+1, 3k+2. There are 2 choices for each of these forms (3k, 3k+1, 3k+2). Thus, there are 222 = 8 ways to choose r1, r2, r3 such that ω^r1 + ω^r2 + ω^r3 = 0.
There are a total of 6^3 = 216 possible outcomes when throwing a fair die three times.
Therefore, the probability is 8/216 = 1/27
This is approximately 0.037
Let's reconsider the calculation:
There are 2 choices for each of 3k, 3k+1, 3k+2 (i.e., 2 choices for each type). So there are 222 = 8 ways to choose r1, r2, r3 such that they satisfy the condition ω^r1 + ω^r2 + ω^r3 = 0. The total number of possible outcomes is 6^3 = 216. Therefore, the probability is 8/216 = 1/27 ≈ 0.037. None of the options match this exactly. There must be a mistake in the question or options.
However, if we consider the permutations, then we have 3! ways to arrange the chosen values. Therefore we have 3! * 2 * 2 * 2 = 6 * 8 = 48. This also is incorrect.
Let's try a different approach. There are 2 choices for each residue class (mod 3). There are 3! = 6 ways to arrange these. So we have 6 * 2 * 2 * 2 = 48 ways. Probability is 48/216 = 2/9. Still not an option.
Required probability = (3! × 2C1 × 2C1 × 2C1) / (6 × 6 × 6) = (6 × 8) / 216 = 48/216 = 2/9
This is still not one of the given options. There appears to be an error in the question or the provided options.