If the points (5, 2, 4), (6, -1, 2) and (8, -7, k) are collinear, then k =

Options

A. -2
B. 2
C. -10
D. 10

Answer

To determine the value of \( k \) such that the points \( A(5, 2, 4) \), \( B(6, -1, 2) \), and \( C(8, -7, k) \) are collinear, we can use the concept that the area of the triangle formed by these three points must be zero.

### Step-by-Step Solution:

1. **Identify the Points**:
We have three points:
- \( A(5, 2, 4) \)
- \( B(6, -1, 2) \)
- \( C(8, -7, k) \)

2. **Use the Area Formula**:
The area of a triangle formed by three points \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) in three-dimensional space can be calculated using the determinant:
\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \right| \]
For our points, this becomes:
\[ \text{Area} = \frac{1}{2} \begin{vmatrix} 5 & 2 & 4 \\ 6 & -1 & 2 \\ 8 & -7 & k \end{vmatrix} \]

3. **Calculate the Determinant**:
We need to compute the determinant:
\[ \begin{vmatrix} 5 & 2 & 4 \\ 6 & -1 & 2 \\ 8 & -7 & k \end{vmatrix} \]
Using the formula for the determinant of a 3x3 matrix:
\[ = 5 \begin{vmatrix} -1 & 2 \\ -7 & k \end{vmatrix} - 2 \begin{vmatrix} 6 & 2 \\ 8 & k \end{vmatrix} + 4 \begin{vmatrix} 6 & -1 \\ 8 & -7 \end{vmatrix} \]

Now, calculate each of the 2x2 determinants:
- \( \begin{vmatrix} -1 & 2 \\ -7 & k \end{vmatrix} = (-1)k - (2)(-7) = -k + 14 \)
- \( \begin{vmatrix} 6 & 2 \\ 8 & k \end{vmatrix} = (6)(k) - (2)(8) = 6k - 16 \)
- \( \begin{vmatrix} 6 & -1 \\ 8 & -7 \end{vmatrix} = (6)(-7) - (-1)(8) = -42 + 8 = -34 \)

Substitute these back into the determinant:
\[ = 5(-k + 14) - 2(6k - 16) + 4(-34) \]
Simplifying this:
\[ = -5k + 70 - 12k + 32 - 136 \]
Combine like terms:
\[ = -17k + 70 + 32 - 136 = -17k - 34 \]

4. **Set the Determinant to Zero**:
Since the points are collinear, the area must be zero:
\[ -17k - 34 = 0 \]

5. **Solve for \( k \)**:
Rearranging gives:
\[ -17k = 34 \implies k = \frac{34}{-17} = -2 \]

### Final Answer:
Thus, the value of \( k \) is \( \boxed{-2} \).