1. **Problem Statement:** We are given a triangle with vertices A, B, and C. The angle at vertex B is $120^\circ$. The sides opposite vertices A and C are both 12 units long, and we need to find the length of side $b$ opposite vertex B. 2. **Relevant Formula:** To find the length of a side in a triangle when we know two sides and the included angle, we use the Law of Cosines: $$b^2 = a^2 + c^2 - 2ac \cos(B)$$ where $a$ and $c$ are the known sides, and $B$ is the included angle. 3. **Substitute Known Values:** Here, $a = 12$, $c = 12$, and $B = 120^\circ$. $$b^2 = 12^2 + 12^2 - 2 \times 12 \times 12 \times \cos(120^\circ)$$ 4. **Calculate Cosine:** Recall that $\cos(120^\circ) = -\frac{1}{2}$. 5. **Simplify:** $$b^2 = 144 + 144 - 2 \times 12 \times 12 \times \left(-\frac{1}{2}\right)$$ $$b^2 = 288 + 144$$ $$b^2 = 432$$ 6. **Find $b$:** $$b = \sqrt{432} = \sqrt{144 \times 3} = 12\sqrt{3}$$ 7. **Answer:** The length of side $b$ is $12\sqrt{3}$. This matches the third option provided.