If `veca = hati - hatj + hatk, vecb= 2 hati + hatj -hatk and vec c = lamda hati -hatj+ lamda hatk` are coplanar, find the value of `lamda.`

Answer

To solve the problem of finding the value of \( \lambda \) for which the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are coplanar, we will use the concept of the scalar triple product. The scalar triple product of three vectors is zero if and only if the vectors are coplanar.

### Step-by-Step Solution:

1. **Identify the Vectors**:
We have the following vectors:
\[ \vec{a} = \hat{i} - \hat{j} + \hat{k} \]
\[ \vec{b} = 2\hat{i} + \hat{j} - \hat{k} \]
\[ \vec{c} = \lambda \hat{i} - \hat{j} + \lambda \hat{k} \]

2. **Set Up the Scalar Triple Product**:
The scalar triple product can be calculated using the determinant of a matrix formed by the components of the vectors:
\[ \text{Scalar Triple Product} = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{vmatrix} \]

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

Now, calculating each of the 2x2 determinants:
- For the first determinant:
\[ = 1(\lambda - 1) = \lambda - 1 \]
- For the second determinant:
\[ = 2\lambda - (-1)(\lambda) = 2\lambda + \lambda = 3\lambda \]
- For the third determinant:
\[ = -2 - \lambda = -2 - \lambda \]

Putting it all together:
\[ \text{Determinant} = (\lambda - 1) + 3\lambda + (-2 - \lambda) \]
Simplifying:
\[ = \lambda - 1 + 3\lambda - 2 - \lambda = 3\lambda - 3 \]

4. **Set the Determinant to Zero**:
Since the vectors are coplanar, we set the determinant equal to zero:
\[ 3\lambda - 3 = 0 \]

5. **Solve for \( \lambda \)**:
\[ 3\lambda = 3 \implies \lambda = 1 \]

### Final Answer:
The value of \( \lambda \) is:
\[ \lambda = 1 \]