The vectors `lamdahat(i)+hat(j)+2hat(k),hat(i)+lamdahat(j)-hat(k)` and `2hat(i)-hat(j)+lamda hat(k)` are coplanar, if

Options

A. `lamda=-2`
B. `lamda=0`
C. `lamda=1`
D. `lamda=-1`

Answer

To determine the value of \( \lambda \) for which the vectors \( \lambda \hat{i} + \hat{j} + 2 \hat{k} \), \( \hat{i} + \lambda \hat{j} - \hat{k} \), and \( 2 \hat{i} - \hat{j} + \lambda \hat{k} \) are coplanar, we can use the condition that the scalar triple product of the vectors must be zero.

### Step-by-step Solution:

1. **Identify the Vectors**:
Let:
\[ \mathbf{A} = \lambda \hat{i} + \hat{j} + 2 \hat{k} \]
\[ \mathbf{B} = \hat{i} + \lambda \hat{j} - \hat{k} \]
\[ \mathbf{C} = 2 \hat{i} - \hat{j} + \lambda \hat{k} \]

2. **Set Up the Determinant**:
The vectors are coplanar if the determinant of the matrix formed by their components is zero:
\[ \begin{vmatrix} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
We will compute the determinant using the formula for a \( 3 \times 3 \) matrix:
\[ D = \lambda \begin{vmatrix} \lambda & -1 \\ -1 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 2 & \lambda \end{vmatrix} + 2 \begin{vmatrix} 1 & \lambda \\ 2 & -1 \end{vmatrix} \]

- First, calculate the smaller determinants:
\[ \begin{vmatrix} \lambda & -1 \\ -1 & \lambda \end{vmatrix} = \lambda^2 - 1 \]
\[ \begin{vmatrix} 1 & -1 \\ 2 & \lambda \end{vmatrix} = \lambda - 2 \]
\[ \begin{vmatrix} 1 & \lambda \\ 2 & -1 \end{vmatrix} = -1 - 2\lambda = -1 - 2\lambda \]

- Substitute these back into the determinant:
\[ D = \lambda(\lambda^2 - 1) - 1(\lambda - 2) + 2(-1 - 2\lambda) \]
\[ = \lambda^3 - \lambda - \lambda + 2 - 2 - 4\lambda \]
\[ = \lambda^3 - 6\lambda \]

4. **Set the Determinant to Zero**:
For coplanarity, we set the determinant equal to zero:
\[ \lambda^3 - 6\lambda = 0 \]
Factor out \( \lambda \):
\[ \lambda(\lambda^2 - 6) = 0 \]

5. **Solve for \( \lambda \)**:
This gives us:
\[ \lambda = 0 \quad \text{or} \quad \lambda^2 - 6 = 0 \]
Solving \( \lambda^2 - 6 = 0 \):
\[ \lambda^2 = 6 \implies \lambda = \sqrt{6} \quad \text{or} \quad \lambda = -\sqrt{6} \]

### Final Answer:
The vectors are coplanar if:
\[ \lambda = 0, \quad \lambda = \sqrt{6}, \quad \text{or} \quad \lambda = -\sqrt{6} \]