1. **Find the area of triangle ABC with vertices A(18, -2, -12), B(14, 4, -10), and C(7, 6, -6).** 2. The area of a triangle with vertices in 3D can be found using the formula: $$\text{Area} = \frac{1}{2} \| \vec{AB} \times \vec{AC} \|$$ where $\vec{AB}$ and $\vec{AC}$ are vectors from vertex A to B and A to C respectively. 3. Calculate vectors: $$\vec{AB} = B - A = (14 - 18, 4 - (-2), -10 - (-12)) = (-4, 6, 2)$$ $$\vec{AC} = C - A = (7 - 18, 6 - (-2), -6 - (-12)) = (-11, 8, 6)$$ 4. Compute the cross product $\vec{AB} \times \vec{AC}$: $$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 6 & 2 \\ -11 & 8 & 6 \end{vmatrix} = \mathbf{i}(6 \times 6 - 2 \times 8) - \mathbf{j}(-4 \times 6 - 2 \times -11) + \mathbf{k}(-4 \times 8 - 6 \times -11)$$ $$= \mathbf{i}(36 - 16) - \mathbf{j}(-24 + 22) + \mathbf{k}(-32 + 66) = \mathbf{i}(20) - \mathbf{j}(-2) + \mathbf{k}(34) = (20, 2, 34)$$ 5. Find the magnitude of the cross product: $$\| \vec{AB} \times \vec{AC} \| = \sqrt{20^2 + 2^2 + 34^2} = \sqrt{400 + 4 + 1156} = \sqrt{1560} = 2\sqrt{390}$$ 6. Calculate the area: $$\text{Area} = \frac{1}{2} \times 2\sqrt{390} = \sqrt{390} \approx 19.75$$ **Final answer:** The area of triangle ABC is approximately $19.75$ square units.