1. **Stating the problem:** We are given three points in 3D space: $A(-2,1,6)$, $B(2,4,5)$, and $C(-1,-2,1)$. We need to show that these points form a right triangle and then find the area of the triangle. 2. **Using vectors to check for a right angle:** We find the vectors representing two sides of the triangle sharing a vertex, for example, vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$: $$\overrightarrow{AB} = B - A = (2 - (-2), 4 - 1, 5 - 6) = (4, 3, -1)$$ $$\overrightarrow{AC} = C - A = (-1 - (-2), -2 - 1, 1 - 6) = (1, -3, -5)$$ 3. **Check if $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are perpendicular:** Two vectors are perpendicular if their dot product is zero: $$\overrightarrow{AB} \cdot \overrightarrow{AC} = 4 \times 1 + 3 \times (-3) + (-1) \times (-5) = 4 - 9 + 5 = 0$$ Since the dot product is zero, $\overrightarrow{AB}$ is perpendicular to $\overrightarrow{AC}$, so angle $A$ is a right angle. 4. **Finding the area of the triangle:** The area of a triangle is half the magnitude of the cross product of two adjacent side vectors: $$\text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right|$$ Calculate the cross product: $$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & -1 \\ 1 & -3 & -5 \end{vmatrix} = \mathbf{i}(3 \times -5 - (-1) \times -3) - \mathbf{j}(4 \times -5 - (-1) \times 1) + \mathbf{k}(4 \times -3 - 3 \times 1)$$ $$= \mathbf{i}(-15 - 3) - \mathbf{j}(-20 - (-1)) + \mathbf{k}(-12 - 3) = \mathbf{i}(-18) - \mathbf{j}(-19) + \mathbf{k}(-15) = (-18, 19, -15)$$ Calculate the magnitude: $$\left| \overrightarrow{AB} \times \overrightarrow{AC} \right| = \sqrt{(-18)^2 + 19^2 + (-15)^2} = \sqrt{324 + 361 + 225} = \sqrt{910}$$ 5. **Final area:** $$\text{Area} = \frac{1}{2} \sqrt{910} = \frac{\sqrt{910}}{2}$$ This is the area of the right triangle formed by points $A$, $B$, and $C$.