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

Options

A. 4
B. 2
C. -2
D. -4

Answer

To find the value of \( \lambda \) such that the points \( A(6, -1, 2) \), \( B(8, -7, \lambda) \), and \( C(5, 2, 4) \) are collinear, we can use the concept of the area of a triangle formed by these points. If the points are collinear, the area of the triangle they form will be zero.

### Step-by-step Solution:

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

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

3. **Calculate Each Minor Determinant**:
- For the first minor:
\[ \begin{vmatrix} -7 & \lambda \\ 2 & 4 \end{vmatrix} = (-7)(4) - (2)(\lambda) = -28 - 2\lambda \]
- For the second minor:
\[ \begin{vmatrix} 8 & \lambda \\ 5 & 4 \end{vmatrix} = (8)(4) - (5)(\lambda) = 32 - 5\lambda \]
- For the third minor:
\[ \begin{vmatrix} 8 & -7 \\ 5 & 2 \end{vmatrix} = (8)(2) - (5)(-7) = 16 + 35 = 51 \]

4. **Substituting Back into the Determinant**:
Now substitute back into the determinant:
\[ = 6(-28 - 2\lambda) + (32 - 5\lambda) + 2(51) \]
Simplifying this:
\[ = -168 - 12\lambda + 32 - 5\lambda + 102 \]
Combine like terms:
\[ = (-168 + 32 + 102) + (-12\lambda - 5\lambda) = -34 - 17\lambda \]

5. **Set the Area to Zero**:
Since the area must be zero for collinearity:
\[ \frac{1}{2} |-34 - 17\lambda| = 0 \]
This implies:
\[ -34 - 17\lambda = 0 \]

6. **Solve for \( \lambda \)**:
Rearranging gives:
\[ 17\lambda = -34 \implies \lambda = -2 \]

### Final Answer:
\[ \lambda = -2 \]