The value of `lamda` for which the four points `2hati+3hatj-hatk,hati+2hatj+3hatk,3hati+4hatj-2hatk` and `hati-lamdahatj+6hatk` are coplanar.

Options

A. 8
B. 0
C. `-2`
D. 6

Explanation

To find the value of `lambda` for which the four points \( P_1(2\hat{i} + 3\hat{j} - \hat{k}) \), \( P_2(\hat{i} + 2\hat{j} + 3\hat{k}) \), \( P_3(3\hat{i} + 4\hat{j} - 2\hat{k}) \), and \( P_4(\hat{i} - \lambda \hat{j} + 6\hat{k}) \) are coplanar, we can follow these steps:

### Step 1: Define the position vectors
Let:
- \( \vec{A} = 2\hat{i} + 3\hat{j} - \hat{k} \)
- \( \vec{B} = \hat{i} + 2\hat{j} + 3\hat{k} \)
- \( \vec{C} = 3\hat{i} + 4\hat{j} - 2\hat{k} \)
- \( \vec{D} = \hat{i} - \lambda \hat{j} + 6\hat{k} \)

### Step 2: Find the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \)
- \( \vec{AB} = \vec{B} - \vec{A} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k}) = -\hat{i} - \hat{j} + 4\hat{k} \)
- \( \vec{AC} = \vec{C} - \vec{A} = (3\hat{i} + 4\hat{j} - 2\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k}) = \hat{i} + \hat{j} - k \)
- \( \vec{AD} = \vec{D} - \vec{A} = (\hat{i} - \lambda \hat{j} + 6\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k}) = -\hat{i} - (\lambda + 3)\hat{j} + 7\hat{k} \)

### Step 3: Set up the determinant for coplanarity
The points are coplanar if the determinant of the matrix formed by the vectors \( \vec{AB} \), \( \vec{AC} \), and \( \vec{AD} \) is equal to zero:

\[ \begin{vmatrix} -1 & -1 & 4 \\ 1 & 1 & -1 \\ -1 & -(\lambda + 3) & 7 \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant:

\[ D = -1 \begin{vmatrix} 1 & -1 \\ -(\lambda + 3) & 7 \end{vmatrix} + 1 \begin{vmatrix} -1 & 4 \\ -1 & 7 \end{vmatrix} + 4 \begin{vmatrix} -1 & -1 \\ -1 & -(\lambda + 3) \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} 1 & -1 \\ -(\lambda + 3) & 7 \end{vmatrix} = 1 \cdot 7 - (-1)(-\lambda - 3) = 7 - \lambda - 3 = 4 - \lambda \)

2. \( \begin{vmatrix} -1 & 4 \\ -1 & 7 \end{vmatrix} = (-1)(7) - (4)(-1) = -7 + 4 = -3 \)

3. \( \begin{vmatrix} -1 & -1 \\ -1 & -(\lambda + 3) \end{vmatrix} = (-1)(-\lambda - 3) - (-1)(-1) = \lambda + 3 - 1 = \lambda + 2 \)

Putting it all together:

\[ D = -1(4 - \lambda) + 1(-3) + 4(\lambda + 2) \]
\[ = -4 + \lambda - 3 + 4\lambda + 8 \]
\[ = 5\lambda + 1 \]

### Step 5: Set the determinant to zero
Setting the determinant equal to zero for coplanarity:

\[ 5\lambda + 1 = 0 \]

### Step 6: Solve for \( \lambda \)
\[ 5\lambda = -1 \implies \lambda = -\frac{1}{5} \]

### Final Answer
Thus, the value of \( \lambda \) for which the four points are coplanar is:

\[ \lambda = -\frac{1}{5} \]