For what value of `'lambda'` are the following vectors coplanar ?
`vec(a)=hat(i)-hat(j)+hat(k), vec(b) = 3hat(i)+hat(j)+2hat(k)` and `vec(c )=hat(i)+lambda hat(j)-3hat(k)`

Answer

To determine the value of \(\lambda\) for which the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar, we can use the condition that the scalar triple product of the vectors must be zero. This can be represented using the determinant of a matrix formed by the coefficients of the vectors.

### Step-by-Step Solution:

1. **Identify the vectors**:
\[ \vec{a} = \hat{i} - \hat{j} + \hat{k} \quad \Rightarrow \quad (1, -1, 1) \]
\[ \vec{b} = 3\hat{i} + \hat{j} + 2\hat{k} \quad \Rightarrow \quad (3, 1, 2) \]
\[ \vec{c} = \hat{i} + \lambda \hat{j} - 3\hat{k} \quad \Rightarrow \quad (1, \lambda, -3) \]

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:
\[ = 1 \cdot \begin{vmatrix} 1 & 2 \\ \lambda & -3 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 3 & 2 \\ 1 & -3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 1 \\ 1 & \lambda \end{vmatrix} \]

Calculating each of the 2x2 determinants:
- First determinant:
\[ = 1 \cdot (-3 - 2\lambda) = -3 - 2\lambda \]
- Second determinant:
\[ = 3 \cdot (-3) - 2 \cdot 1 = -9 - 2 = -11 \quad \Rightarrow \quad \text{(since we have a negative sign in front)} \quad = +11 \]
- Third determinant:
\[ = 3\lambda - 1 \]

4. **Combine the results**:
Putting it all together:
\[ -3 - 2\lambda + 11 + 3\lambda - 1 = 0 \]
Simplifying:
\[ (3\lambda - 2\lambda) + (-3 + 11 - 1) = 0 \]
\[ \lambda + 7 = 0 \]

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

### Final Answer:
The value of \(\lambda\) for which the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar is \(\lambda = -7\).