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To solve this problem, we’ll break it into two parts:

1. **Find a nonzero vector orthogonal to the plane** containing the points \( P = (0, 0, -3) \), \( Q = (4, 2, 0) \), and \( R = (3, 3, 1) \).
2. **Calculate the area of triangle \( PQR \)**.

### Step 1: Find a Nonzero Vector Orthogonal to the Plane

To find a vector orthogonal to the plane, we can use the cross product of two vectors that lie in the plane. We’ll construct two vectors in the plane from the given points:

1. **Vector \( \overrightarrow{PQ} \):**
   \[
   \overrightarrow{PQ} = Q - P = (4 - 0, 2 - 0, 0 - (-3)) = (4, 2, 3)
   \]

2. **Vector \( \overrightarrow{PR} \):**
   \[
   \overrightarrow{PR} = R - P = (3 - 0, 3 - 0, 1 - (-3)) = (3, 3, 4)
   \]

Now, calculate the cross product \( \overrightarrow{PQ} \times \overrightarrow{PR} \) to find a vector orthogonal to the plane:

\[
\overrightarrow{PQ} \times \overrightarrow{PR} = 
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
4 & 2 & 3 \\
3 & 3 & 4 \\
\end{vmatrix}
\]

Expanding this determinant:

\[
= \hat{i}(2 \cdot 4 - 3 \cdot 3) - \hat{j}(4 \cdot 4 - 3 \cdot 3) + \hat{k}(4 \cdot 3 - 2 \cdot 3)
\]
\[
= \hat{i}(8 - 9) - \hat{j}(16 - 9) + \hat{k}(12 - 6)
\]
\[
= -\hat{i} - 7\hat{j} + 6\hat{k}
\]

So, a nonzero vector orthogonal to the plane is:
\[
\vec{n} = (-1, -7, 6)
\]

### Step 2: Find the Area of Triangle \( PQR \)

The area of triangle \( PQR \) is given by:
\[
\text{Area} = \frac{1}{2} \|\overrightarrow{PQ} \times \overrightarrow{PR}\|
\]

First, find the magnitude of \( \overrightarrow{PQ} \times \overrightarrow{PR} \):
\[
\|\overrightarrow{PQ} \times \overrightarrow{PR}\| = \sqrt{(-1)^2 + (-7)^2 + 6^2}
\]
\[
= \sqrt{1 + 49 + 36} = \sqrt{86}
\]

Thus, the area of triangle \( PQR \) is:
\[
\text{Area} = \frac{1}{2} \sqrt{86}
\]

### Final Answers

1. **A nonzero vector orthogonal to the plane**: \( \vec{n} = (-1, -7, 6) \).
2. **The area of triangle \( PQR \)**: \( \frac{1}{2} \sqrt{86} \) (approximately 4.64 square units).

Would you like further details on any specific step? 

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Here are 5 additional questions for practice:
1. How would you find the equation of the plane containing points \( P \), \( Q \), and \( R \)?
2. What would happen if two of the points \( P \), \( Q \), and \( R \) were collinear?
3. How do you interpret the cross product in terms of geometry?
4. Can a different orthogonal vector be found using scalar multiples of \( \vec{n} \)?
5. How would you verify that \( \vec{n} \) is indeed orthogonal to the plane?

**Tip:** Cross products are useful for finding perpendicular vectors in 3D geometry; remember that the magnitude of the cross product also gives the area of the parallelogram formed by the two vectors.

Find a nonzero vector orthogonal to the plane through points P = (0, 0, -3), Q = (4, 2, 0), R = (3, 3, 1). Then find the area of triangle PQR.

Solve for a nonzero vector orthogonal to the plane defined by points P, Q, and R in 3D space, and calculate the area of triangle PQR using cross product. This problem covers concepts in vector math and 3D geometry, including cross product calculation and triangle area in 3D space.

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