Find `'lambda'` . If the vectors `lambdahati+hatj+2hatk, 2hati-hatj+lambdahatk`, and `hati+lambdahatj-hatk` are coplanar.

Answer

To find the value of \( \lambda \) such that the vectors \( \lambda \hat{i} + \hat{j} + 2 \hat{k} \), \( 2 \hat{i} - \hat{j} + \lambda \hat{k} \), and \( \hat{i} + \lambda \hat{j} - \hat{k} \) are coplanar, we can use the condition that the scalar triple product of the vectors is zero. This can be expressed using the determinant of a matrix formed by the components of the vectors.

### Step-by-Step Solution:

1. **Write the vectors in component form:**
- Let \( \mathbf{A} = \lambda \hat{i} + \hat{j} + 2 \hat{k} \)
- Let \( \mathbf{B} = 2 \hat{i} - \hat{j} + \lambda \hat{k} \)
- Let \( \mathbf{C} = \hat{i} + \lambda \hat{j} - \hat{k} \)

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

3. **Calculate the determinant:**
Expanding the determinant:
\[ = \lambda \begin{vmatrix} -1 & \lambda \\ \lambda & -1 \end{vmatrix} - 1 \begin{vmatrix} 2 & \lambda \\ 1 & -1 \end{vmatrix} + 2 \begin{vmatrix} 2 & -1 \\ 1 & \lambda \end{vmatrix} \]

Now calculating each of these 2x2 determinants:
- First determinant:
\[ = \lambda((-1)(-1) - \lambda^2) = \lambda(1 - \lambda^2) = \lambda - \lambda^3 \]
- Second determinant:
\[ = 2(-1) - \lambda(1) = -2 - \lambda \]
- Third determinant:
\[ = 2\lambda - (-1)(1) = 2\lambda + 1 \]

Putting it all together:
\[ \lambda(1 - \lambda^2) - ( -2 - \lambda) + 2(2\lambda + 1) = 0 \]
Simplifying:
\[ \lambda - \lambda^3 + 2 + \lambda + 4\lambda + 2 = 0 \]
\[ -\lambda^3 + 6\lambda + 4 = 0 \]

4. **Rearranging the equation:**
\[ \lambda^3 - 6\lambda - 4 = 0 \]

5. **Finding the roots of the cubic equation:**
We can use trial and error to find rational roots. Testing \( \lambda = -2 \):
\[ (-2)^3 - 6(-2) - 4 = -8 + 12 - 4 = 0 \]
Thus, \( \lambda = -2 \) is a root.

6. **Conclusion:**
The value of \( \lambda \) for which the given vectors are coplanar is:
\[ \lambda = -2 \]