If vectors `veca =hati +2hatj -hatk, vecb = 2hati -hatj +hatk and vecc = lamdahati +hatj +2hatk` are coplanar, then find the value of `(lamda -4)`.

Explanation

To find the value of \( \lambda - 4 \) given that the vectors \( \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \), \( \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \), and \( \vec{c} = \lambda\hat{i} + \hat{j} + 2\hat{k} \) are coplanar, we can follow these steps:

### Step 1: Understand the condition for coplanarity
Vectors are coplanar if the scalar triple product of the vectors is zero. This can be expressed using the determinant of a matrix formed by the vectors.

### Step 2: Set up the determinant
We will set up the determinant using the components of the vectors:
\[ \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 1 \\ \lambda & 1 & 2 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
We will calculate the determinant:
\[ \text{Det} = 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\ \lambda & 2 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & -1 \\ \lambda & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 1 \\ 1 & 2 \end{vmatrix} = (-1)(2) - (1)(1) = -2 - 1 = -3 \)

2. \( \begin{vmatrix} 2 & 1 \\ \lambda & 2 \end{vmatrix} = (2)(2) - (1)(\lambda) = 4 - \lambda \)

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

Putting these back into the determinant:
\[ \text{Det} = 1(-3) - 2(4 - \lambda) + (2 + \lambda) = -3 - 8 + 2\lambda + 2 + \lambda \]
\[ = -9 + 3\lambda \]

### Step 4: Set the determinant to zero
Since the vectors are coplanar:
\[ -9 + 3\lambda = 0 \]

### Step 5: Solve for \( \lambda \)
\[ 3\lambda = 9 \implies \lambda = 3 \]

### Step 6: Find \( \lambda - 4 \)
Now we need to find \( \lambda - 4 \):
\[ \lambda - 4 = 3 - 4 = -1 \]

### Final Answer
The value of \( \lambda - 4 \) is \( -1 \).

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