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To find the area of a triangle in 3D space with vertices \( P(-2, 6, 1) \), \( Q(5, 5, 4) \), and \( R(-3, -3, -5) \), we can use the formula based on the cross product of two vectors formed from these points.

1. **Step 1:** Calculate vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \):
   - \( \overrightarrow{PQ} = Q - P = (5 - (-2), 5 - 6, 4 - 1) = (7, -1, 3) \)
   - \( \overrightarrow{PR} = R - P = (-3 - (-2), -3 - 6, -5 - 1) = (-1, -9, -6) \)

2. **Step 2:** Find the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \):
   \[
   \overrightarrow{PQ} \times \overrightarrow{PR} = 
   \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   7 & -1 & 3 \\
   -1 & -9 & -6 \\
   \end{vmatrix}
   \]
   Expanding the determinant:
   \[
   \overrightarrow{PQ} \times \overrightarrow{PR} = \mathbf{i}((-1)(-6) - (3)(-9)) - \mathbf{j}((7)(-6) - (3)(-1)) + \mathbf{k}((7)(-9) - (-1)(-1))
   \]
   Simplifying each term:
   \[
   = \mathbf{i}(6 + 27) - \mathbf{j}(-42 + 3) + \mathbf{k}(-63 - 1)
   \]
   \[
   = \mathbf{i}(33) - \mathbf{j}(-39) + \mathbf{k}(-64)
   \]
   \[
   = 33\mathbf{i} + 39\mathbf{j} - 64\mathbf{k}
   \]
   So, \( \overrightarrow{PQ} \times \overrightarrow{PR} = (33, 39, -64) \).

3. **Step 3:** Find the magnitude of the cross product \( |\overrightarrow{PQ} \times \overrightarrow{PR}| \):
   \[
   |\overrightarrow{PQ} \times \overrightarrow{PR}| = \sqrt{33^2 + 39^2 + (-64)^2}
   \]
   \[
   = \sqrt{1089 + 1521 + 4096} = \sqrt{6706}
   \]

4. **Step 4:** Calculate the area of the triangle:
   The area \( A \) of the triangle is half the magnitude of the cross product:
   \[
   A = \frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}|
   \]
   \[
   = \frac{1}{2} \sqrt{6706} \approx \frac{1}{2} \times 81.89 \approx 40.945
   \]

### Answer:
The area of the triangle \( PQR \) is approximately **40.945 square units**.

Would you like further details or have any questions?

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Here are five related questions:
1. How do we calculate the cross product of two vectors in three dimensions?
2. Why is the cross product useful for finding the area of a triangle in 3D space?
3. Can we use this method for calculating areas of triangles in 2D?
4. What is the significance of the direction of the cross product in 3D geometry?
5. How does the distance formula relate to vector magnitudes?

**Tip:** To verify your cross product calculation, remember that each component involves a pairwise determinant of the vector’s components.

Find the area of a triangle PQR, where P = (-2, 6, 1), Q = (5, 5, 4), and R = (-3, -3, -5).

Learn how to calculate the area of a triangle with vertices P = (-2, 6, 1), Q = (5, 5, 4), and R = (-3, -3, -5) in 3D space. This solution covers vector calculations, the cross product, and area determination.

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