To solve the problems related to the points $P(2, -2, 1)$, $Q(1, 0, -2)$, and $R(0, 1, 2)$, we will go through the following steps:

Part (a): Finding the Area of the Triangle Determined by Points $P$, $Q$, and $R$

The area of the triangle can be found using the formula: $\text{Area} = \frac{1}{2} \left\| \overrightarrow{PQ} \times \overrightarrow{PR} \right\|$ where: $\overrightarrow{PQ} = Q - P = (1-2, 0-(-2), -2-1) = (-1, 2, -3)$ $\overrightarrow{PR} = R - P = (0-2, 1-(-2), 2-1) = (-2, 3, 1)$

Now, compute the cross product $\overrightarrow{PQ} \times \overrightarrow{PR}$:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & -3 \\ -2 & 3 & 1 \\ \end{vmatrix}$$ Expanding this determinant: $$\overrightarrow{PQ} \times \overrightarrow{PR} = \mathbf{i}(2 \times 1 - (-3) \times 3) - \mathbf{j}((-1) \times 1 - (-3) \times (-2)) + \mathbf{k}((-1) \times 3 - 2 \times (-2))$$ $$= \mathbf{i}(2 + 9) - \mathbf{j}(-1 - 6) + \mathbf{k}(-3 + 4)$$ $$= \mathbf{i}(11) - \mathbf{j}(-7) + \mathbf{k}(1)$$ $$\overrightarrow{PQ} \times \overrightarrow{PR} = (11, 7, 1)$$ The magnitude of the cross product is: $$\left\| \overrightarrow{PQ} \times \overrightarrow{PR} \right\| = \sqrt{11^2 + 7^2 + 1^2} = \sqrt{121 + 49 + 1} = \sqrt{171}$$ Therefore, the area of the triangle is: $$\text{Area} = \frac{1}{2} \times \sqrt{171} \approx \frac{1}{2} \times 13.08 = 6.54$$ ### Part (b): Finding a Unit Vector Perpendicular to Plane $$PQR$$ A vector perpendicular to the plane can be found by normalizing the cross product $$\overrightarrow{PQ} \times \overrightarrow{PR}$$. The unit vector is: $$\text{Unit vector} = \frac{\overrightarrow{PQ} \times \overrightarrow{PR}}{\left\| \overrightarrow{PQ} \times \overrightarrow{PR} \right\|} = \frac{(11, 7, 1)}{\sqrt{171}}$$ $$= \left( \frac{11}{\sqrt{171}}, \frac{7}{\sqrt{171}}, \frac{1}{\sqrt{171}} \right)$$ So, the unit vector perpendicular to the plane $$PQR$$ is: $$\left( \frac{11}{\sqrt{171}}, \frac{7}{\sqrt{171}}, \frac{1}{\sqrt{171}} \right)$$ ### Summary: - **Area of the triangle:** $$6.54$$ square units. - **Unit vector perpendicular to the plane:** $$\left( \frac{11}{\sqrt{171}}, \frac{7}{\sqrt{171}}, \frac{1}{\sqrt{171}} \right)$$. Would you like more details on any part of the solution? Here are some related questions: 1. How is the cross product calculated for vectors in three dimensions? 2. What are the properties of the cross product in vector analysis? 3. How can the magnitude of a vector be determined? 4. How does the area of a triangle relate to the cross product? 5. What are the applications of unit vectors in physics and engineering? **Tip:** Remember, the cross product of two vectors gives a vector that is perpendicular to both, which is crucial in determining the normal to a plane.