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

Explanation

To determine the value of \( \lambda \) for which 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 use the condition that the scalar triple product of the vectors must equal zero. This can be represented using a determinant.

### Step-by-Step Solution:

1. **Write the vectors in component form**:
\[ \vec{a} = (1, 2, -1), \quad \vec{b} = (2, -1, 1), \quad \vec{c} = (\lambda, 1, 2) \]

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

3. **Calculate the determinant**:
Using the formula for the determinant of a 3x3 matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where:
- \( a = 1, b = 2, c = -1 \)
- \( d = 2, e = -1, f = 1 \)
- \( g = \lambda, h = 1, i = 2 \)

The determinant can be calculated as follows:
\[ = 1((-1)(2) - (1)(1)) - 2((2)(2) - (1)(\lambda)) - 1((2)(1) - (-1)(\lambda)) \]
\[ = 1(-2 - 1) - 2(4 - \lambda) - 1(2 + \lambda) \]
\[ = 1(-3) - 2(4 - \lambda) - (2 + \lambda) \]
\[ = -3 - (8 - 2\lambda) - (2 + \lambda) \]
\[ = -3 - 8 + 2\lambda - 2 - \lambda \]
\[ = -13 + \lambda \]

4. **Set the determinant equal to zero**:
\[ -13 + \lambda = 0 \]

5. **Solve for \( \lambda \)**:
\[ \lambda = 13 \]

### Final Answer:
The value of \( \lambda \) is \( 13 \).