If `veca=2hati+lambda hatj+3hatk, vecb=3hati+3hatj+5hatk, vec c=lambda hati+2hatj+2hatk` are linearly dependent vectors, then the number of possible values of `lambda` is :

Options

A. 0
B. 1
C. 2
D. More than 2

Answer

To determine the number of possible values of \( \lambda \) for which the vectors \( \vec{a} = 2\hat{i} + \lambda \hat{j} + 3\hat{k} \), \( \vec{b} = 3\hat{i} + 3\hat{j} + 5\hat{k} \), and \( \vec{c} = \lambda \hat{i} + 2\hat{j} + 2\hat{k} \) are linearly dependent, we need to set up the determinant of the matrix formed by these vectors and solve for \( \lambda \).

### Step-by-Step Solution:

1. **Form the Matrix**: We can represent the vectors as a matrix:
\[ \begin{bmatrix} 2 & \lambda & 3 \\ 3 & 3 & 5 \\ \lambda & 2 & 2 \end{bmatrix} \]

2. **Set Up the Determinant**: For the vectors to be linearly dependent, the determinant of this matrix must equal zero:
\[ \text{det} \begin{bmatrix} 2 & \lambda & 3 \\ 3 & 3 & 5 \\ \lambda & 2 & 2 \end{bmatrix} = 0 \]

3. **Calculate the Determinant**: We can expand the determinant using the first row:
\[ = 2 \begin{vmatrix} 3 & 5 \\ 2 & 2 \end{vmatrix} - \lambda \begin{vmatrix} 3 & 5 \\ \lambda & 2 \end{vmatrix} + 3 \begin{vmatrix} 3 & 3 \\ \lambda & 2 \end{vmatrix} \]

4. **Calculate Each Minor**:
- First minor:
\[ \begin{vmatrix} 3 & 5 \\ 2 & 2 \end{vmatrix} = (3 \cdot 2) - (5 \cdot 2) = 6 - 10 = -4 \]
- Second minor:
\[ \begin{vmatrix} 3 & 5 \\ \lambda & 2 \end{vmatrix} = (3 \cdot 2) - (5 \cdot \lambda) = 6 - 5\lambda \]
- Third minor:
\[ \begin{vmatrix} 3 & 3 \\ \lambda & 2 \end{vmatrix} = (3 \cdot 2) - (3 \cdot \lambda) = 6 - 3\lambda \]

5. **Substitute Back**:
\[ = 2(-4) - \lambda(6 - 5\lambda) + 3(6 - 3\lambda) \]
\[ = -8 - 6\lambda + 5\lambda^2 + 18 - 9\lambda \]
\[ = 5\lambda^2 - 15\lambda + 10 \]

6. **Set the Determinant to Zero**:
\[ 5\lambda^2 - 15\lambda + 10 = 0 \]

7. **Simplify the Equation**:
Divide the entire equation by 5:
\[ \lambda^2 - 3\lambda + 2 = 0 \]

8. **Factor the Quadratic**:
\[ (\lambda - 2)(\lambda - 1) = 0 \]

9. **Find the Values of \( \lambda \)**:
Setting each factor to zero gives:
\[ \lambda - 2 = 0 \quad \Rightarrow \quad \lambda = 2 \]
\[ \lambda - 1 = 0 \quad \Rightarrow \quad \lambda = 1 \]

10. **Conclusion**: The possible values of \( \lambda \) are \( 1 \) and \( 2 \). Therefore, the number of possible values of \( \lambda \) is **2**.