To find the area of the triangle $PQR$ with vertices $P = (0, -2, -3)$, $Q = (-2, 2, 1)$, and $R = (0, 6, 6)$, we can use the formula based on the cross product of two vectors formed by the points. The area $A$ of the triangle with vertices $P$, $Q$, and $R$ in 3D space is given by:

$A = \frac{1}{2} \| \vec{PQ} \times \vec{PR} \|$

Step-by-Step Solution:

1. Calculate vectors:

   • $\vec{PQ} = Q - P = (-2 - 0, 2 - (-2), 1 - (-3)) = (-2, 4, 4)$
   • $\vec{PR} = R - P = (0 - 0, 6 - (-2), 6 - (-3)) = (0, 8, 9)$

2. Find the cross product $\vec{PQ} \times \vec{PR}$: $\vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 4 & 4 \\ 0 & 8 & 9 \end{vmatrix}$ Expanding this determinant: $= \mathbf{i} (4 \cdot 9 - 4 \cdot 8) - \mathbf{j} (-2 \cdot 9 - 4 \cdot 0) + \mathbf{k} (-2 \cdot 8 - 4 \cdot 0)$ $= \mathbf{i} (36 - 32) - \mathbf{j} (-18) + \mathbf{k} (-16)$ $= 4\mathbf{i} + 18\mathbf{j} - 16\mathbf{k}$ So, $\vec{PQ} \times \vec{PR} = (4, 18, -16)$.

3. Calculate the magnitude of the cross product: $\| \vec{PQ} \times \vec{PR} \| = \sqrt{4^2 + 18^2 + (-16)^2}$ $= \sqrt{16 + 324 + 256} = \sqrt{596} = 2\sqrt{149}$

4. Find the area of the triangle: $A = \frac{1}{2} \| \vec{PQ} \times \vec{PR} \| = \frac{1}{2} \times 2\sqrt{149} = \sqrt{149}$

Final Answer:

$\text{Area of } \triangle PQR = \sqrt{149}$

Would you like further clarification on any part of this solution?

Here are five additional questions related to this topic:

1. How do you compute the magnitude of a vector in 3D space?
2. What is the significance of the cross product in vector geometry?
3. How can you verify if three points in space form a triangle or are collinear?
4. How do you determine the orientation of a triangle given three vertices?
5. What are some applications of using the cross product to find areas in physics?

Tip: When computing cross products, make sure to follow the correct order of vector entries in the determinant to avoid sign errors.