Probability of Sum of Complex Roots of Unity Being Zero

To find the probability that $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$, we analyze the properties of the complex cube roots of unity.

Since $\omega$ is a non-real complex cube root of unity ($\omega \neq 1$), it satisfies the identity:
$1 + \omega + \omega^2 = 0$

A fair die has 6 outcomes: $\{1, 2, 3, 4, 5, 6\}$. Since $r_1, r_2,$ and $r_3$ are obtained from throwing the die three times, the total number of sample space outcomes is:
$n(S) = 6 \times 6 \times 6 = 216$

We classify the possible values on the die according to their remainder when divided by $3$:

1. $r \equiv 1 \pmod 3 \implies r \in \{1, 4\}$, so $\omega^r = \omega$ ($2$ outcomes)
2. $r \equiv 2 \pmod 3 \implies r \in \{2, 5\}$, so $\omega^r = \omega^2$ ($2$ outcomes)
3. $r \equiv 0 \pmod 3 \implies r \in \{3, 6\}$, so $\omega^r = 1$ ($2$ outcomes)

For the sum $\omega^{r_1} + \omega^{r_2} + \omega^{r_3}$ to equal $0$, the set of values $\{\omega^{r_1}, \omega^{r_2}, \omega^{r_3}\}$ must consist of one $1$, one $\omega$, and one $\omega^2$.

Thus, $\{r_1 \pmod 3, r_2 \pmod 3, r_3 \pmod 3\}$ must be a permutation of $\{0, 1, 2\}$.

1. Number of permutations of remainders $(0, 1, 2)$:
   $3! = 6$

2. Number of choices for $r_1, r_2, r_3$ for a fixed permutation of remainders:
   For each remainder class, there are $2$ possible choices on the die. Therefore, for any fixed permutation (e.g., $r_1 \equiv 1, r_2 \equiv 2, r_3 \equiv 0 \pmod 3$), the number of favorable triples $(r_1, r_2, r_3)$ is:
   $2 \times 2 \times 2 = 8$

3. Total favorable outcomes:
   $n(E) = 6 \times 8 = 48$

The required probability $P(E)$ is given by:
$P(E) = \frac{n(E)}{n(S)} = \frac{48}{216} = \frac{2}{9}$

Hence, the correct option is C.