The vectors `veca=hati-2hatj+3hatk, vecb=-2hati+3hatj-4hatk` and `vec c=hati-3hatj+lambdahatk` are coplanar if the value of `lambda` is

Options

A. 1
B. 2
C. 3
D. 5

Answer

To determine the value of \( \lambda \) for which the vectors \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k} \), and \( \vec{c} = \hat{i} - 3\hat{j} + \lambda\hat{k} \) are coplanar, we will use the property that three vectors are coplanar if the scalar triple product is zero. This can be represented using the determinant of a matrix formed by the vectors.

### Step-by-Step Solution:

1. **Write the vectors**:
\[ \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \\ \vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k} \\ \vec{c} = \hat{i} - 3\hat{j} + \lambda\hat{k} \]

2. **Set up the determinant**:
The vectors are coplanar if the determinant of the matrix formed by their coefficients is zero:
\[ \begin{vmatrix} 1 & -2 & 1 \\ -2 & 3 & -3 \\ 3 & -4 & \lambda \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
Expanding the determinant:
\[ = 1 \cdot \begin{vmatrix} 3 & -3 \\ -4 & \lambda \end{vmatrix} - (-2) \cdot \begin{vmatrix} -2 & -3 \\ 3 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} -2 & 3 \\ 3 & -4 \end{vmatrix} \]

Now, calculating each of the 2x2 determinants:
\[ = 1 \cdot (3\lambda - 12) + 2 \cdot (-2\lambda + 9) + 1 \cdot (8 - 9) \]
\[ = 3\lambda - 12 - 4\lambda + 18 - 1 \]
\[ = (3\lambda - 4\lambda) + (-12 + 18 - 1) \]
\[ = -\lambda + 5 \]

4. **Set the determinant to zero**:
\[ -\lambda + 5 = 0 \]

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

### Conclusion:
The value of \( \lambda \) for which the vectors are coplanar is \( \lambda = 5 \).