Find `'lambda'` if the vectors :
`hat(i)-hat(j)+hat(k), 3hat(i)+hat(j)+2hat(k)` and `hat(i)+lambda hat(j)-3hat(k)` are coplanar.

Answer

To find the value of \( \lambda \) such that the vectors \( \mathbf{A} = \hat{i} - \hat{j} + \hat{k} \), \( \mathbf{B} = 3\hat{i} + \hat{j} + 2\hat{k} \), and \( \mathbf{C} = \hat{i} + \lambda \hat{j} - 3\hat{k} \) are coplanar, we can use the condition that the scalar triple product of the vectors is zero.

### Step-by-Step Solution:

1. **Identify the Vectors**:
\[ \mathbf{A} = \hat{i} - \hat{j} + \hat{k} \quad \text{(Coefficients: 1, -1, 1)} \]
\[ \mathbf{B} = 3\hat{i} + \hat{j} + 2\hat{k} \quad \text{(Coefficients: 3, 1, 2)} \]
\[ \mathbf{C} = \hat{i} + \lambda \hat{j} - 3\hat{k} \quad \text{(Coefficients: 1, \lambda, -3)} \]

2. **Set Up the Determinant**:
The vectors are coplanar if the determinant of the matrix formed by their coefficients is zero:
\[ \begin{vmatrix} 1 & -1 & 1 \\ 3 & 1 & 2 \\ 1 & \lambda & -3 \end{vmatrix} = 0 \]

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

Now calculate each of the 2x2 determinants:
- First determinant:
\[ = 1 \cdot (-3 - 2\lambda) = -3 - 2\lambda \]
- Second determinant:
\[ = 3 \cdot (-3) - 2 \cdot 1 = -9 - 2 = -11 \]
- Third determinant:
\[ = 3\lambda - 1 \]

Putting it all together:
\[ -3 - 2\lambda + 11 + 3\lambda - 1 = 0 \]

4. **Simplify the Equation**:
Combine like terms:
\[ (3\lambda - 2\lambda) + (-3 + 11 - 1) = 0 \]
\[ \lambda + 7 = 0 \]

5. **Solve for \( \lambda \)**:
\[ \lambda = -7 \]

### Final Answer:
The value of \( \lambda \) is \( -7 \).