Let `veca=3hati+hatj , vecb=hati+2hatj+hatk and veca xx (vecb xx vecc)=vecb + lamda vecc, vecb` is non parallel to `vecc`, then value of `lamda` is:

Options

A. 5
B. -5
C. 1
D. -1

Answer

To solve the problem, we need to find the value of \( \lambda \) given the vectors \( \vec{a} \) and \( \vec{b} \), and the equation \( \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c} \).

### Step-by-Step Solution:

1. **Identify the Vectors**:
\[ \vec{a} = 3\hat{i} + \hat{j} \]
\[ \vec{b} = \hat{i} + 2\hat{j} + \hat{k} \]

2. **Use the Vector Triple Product Identity**:
The vector triple product identity states that:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \]
Applying this to our case:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

3. **Set Up the Equation**:
From the problem statement, we have:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c} \]
Thus, we can equate:
\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{b} + \lambda \vec{c} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ (\vec{a} \cdot \vec{c}) \vec{b} - \vec{b} = \lambda \vec{c} + (\vec{a} \cdot \vec{b}) \vec{c} \]
This simplifies to:
\[ ((\vec{a} \cdot \vec{c}) - 1) \vec{b} = (\lambda + \vec{a} \cdot \vec{b}) \vec{c} \]

5. **Finding the Dot Products**:
- Calculate \( \vec{a} \cdot \vec{b} \):
\[ \vec{a} \cdot \vec{b} = (3\hat{i} + \hat{j}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = 3 \cdot 1 + 1 \cdot 2 + 0 = 3 + 2 = 5 \]

6. **Substituting Back**:
Substitute \( \vec{a} \cdot \vec{b} = 5 \):
\[ ((\vec{a} \cdot \vec{c}) - 1) \vec{b} = (\lambda + 5) \vec{c} \]

7. **Non-parallel Condition**:
Since \( \vec{b} \) is non-parallel to \( \vec{c} \), we can equate the coefficients:
\[ (\vec{a} \cdot \vec{c}) - 1 = 0 \quad \text{and} \quad \lambda + 5 = 0 \]

8. **Solving for \( \lambda \)**:
From \( \lambda + 5 = 0 \):
\[ \lambda = -5 \]

### Final Answer:
The value of \( \lambda \) is \( -5 \).