Given three non-collinear points P(3,-2,1), Q(0,4,-3), and R(k,2k,k-1), determine the value of k for which the area of triangle PQR is 15 square units.

To determine the value of \( k \) for which the area of triangle \( PQR \) is 15 square units, we can use the formula for the area of a triangle formed by three points in 3D space. The area \( A \) of triangle \( PQR \) is given by:

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

where \( \vec{PQ} = Q - P \) and \( \vec{PR} = R - P \).

### Step 1: Calculate vectors \( \vec{PQ} \) and \( \vec{PR} \)

1. **Calculate \( \vec{PQ} \)**:
   \[
   \vec{PQ} = Q - P = (0 - 3, 4 - (-2), -3 - 1) = (-3, 6, -4)
   \]

2. **Calculate \( \vec{PR} \)**:
   \[
   \vec{PR} = R - P = (k - 3, 2k - (-2), (k - 1) - 1) = (k - 3, 2k + 2, k - 2)
   \]

### Step 2: Calculate the cross product \( \vec{PQ} \times \vec{PR} \)

The cross product \( \vec{PQ} \times \vec{PR} \) can be computed using the determinant:

\[
\vec{PQ} \times \vec{PR} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-3 & 6 & -4 \\
k-3 & 2k + 2 & k - 2
\end{vmatrix}
\]

Calculating this determinant:

\[
= \hat{i} \left( 6(k-2) - (-4)(2k+2) \right) - \hat{j} \left( -3(k-2) - (-4)(k-3) \right) + \hat{k} \left( -3(2k+2) - 6(k-3) \right)
\]

Expanding each component:

1. **For \( \hat{i} \)**:
   \[
   6(k - 2) + 8k + 8 = 6k - 12 + 8k + 8 = 14k - 4
   \]

2. **For \( \hat{j} \)**:
   \[
   -3(k - 2) + 4(k - 3) = -3k + 6 + 4k - 12 = k - 6
   \]

3. **For \( \hat{k} \)**:
   \[
   -6k - 6 - 6k + 18 = -12k + 12
   \]

So, the cross product is:

\[
\vec{PQ} \times \vec{PR} = (14k - 4, - (k - 6), -12k + 12) = (14k - 4, -k + 6, -12k + 12)
\]

### Step 3: Calculate the magnitude of the cross product

\[
\| \vec{PQ} \times \vec{PR} \| = \sqrt{(14k - 4)^2 + (-k + 6)^2 + (-12k + 12)^2}
\]

Expanding each term:

1. \((14k - 4)^2 = 196k^2 - 112k + 16\)
2. \((-k + 6)^2 = k^2 - 12k + 36\)
3. \((-12k + 12)^2 = 144k^2 - 288k + 144\)

Adding these:

\[
= (196k^2 + k^2 + 144k^2) + (-112k - 12k - 288k) + (16 + 36 + 144)
\]
\[
= 341k^2 - 412k + 196
\]

### Step 4: Set the area to 15 square units

Using the area formula:

\[
\frac{1}{2} \sqrt{341k^2 - 412k + 196} = 15
\]

Multiply by 2:

\[
\sqrt{341k^2 - 412k + 196} = 30
\]

Square both sides:

\[
341k^2 - 412k + 196 = 900
\]

Rearranging gives:

\[
341k^2 - 412k - 704 = 0
\]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 341, b = -412, c = -704 \):

\[
b^2 - 4ac = (-412)^2 - 4(341)(-704) = 169744 + 960176 = 1132920
\]

\[
k = \frac{412 \pm \sqrt{1132920}}{682}
\]

Calculating \( \sqrt{1132920} \approx 1064.67 \):

\[
k = \frac{412 \pm 1064.67}{682}
\]

Calculating the two possible values for \( k \):

1. \( k_1 = \frac{412 + 1064.67}{682} \approx 2.14 \)
2. \( k_2 = \frac{412 - 1064.67}{682} \approx -0.96 \)

### Conclusion

The values of \( k \) for which the area of triangle \( PQR \) is 15 square units are approximately \( k \approx 2.14 \) and \( k \approx -0.96 \).

If you have any questions or need further details, feel free to ask! 

Here are some related questions to consider:
1. How would the area change if the points were collinear?
2. What geometric interpretation can be derived from the area of a triangle in 3D space?
3. Can you derive the coordinates of the centroid of triangle \( PQR \)?
4. How does the formula for the area of a triangle in 3D differ from that in 2D?
5. What would happen to the area if one of the points was moved along the z-axis? 

**Tip:** Always verify your calculations for the area, especially when dealing with determinants and cross products.

Explore how to calculate the area of triangle PQR formed by points P(3,-2,1), Q(0,4,-3), and R(k,2k,k-1) in 3D space, finding values of k for an area of 15 square units. This problem integrates concepts of geometry and vectors, suitable for Grades 10-12.

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