If a,b,c are three non-coplanar vectors, then `3a-7b-4c,3a-2b+c and a+b+lamdac` will be coplanar, if `lamda` is

Options

A. `-1`
B. `1`
C. 3
D. 2

Answer

To determine the value of \( \lambda \) such that the vectors \( 3a - 7b - 4c \), \( 3a - 2b + c \), and \( a + b + \lambda c \) are coplanar, we can use the condition that the scalar triple product of these vectors must be zero.

### Step-by-Step Solution:

1. **Define the Vectors**:
Let:
\[ \alpha = 3a - 7b - 4c \]
\[ \beta = 3a - 2b + c \]
\[ \gamma = a + b + \lambda c \]

2. **Set Up the Scalar Triple Product**:
The vectors \( \alpha, \beta, \gamma \) are coplanar if:
\[ \alpha \cdot (\beta \times \gamma) = 0 \]
This can be expressed using the determinant:
\[ \begin{vmatrix} 3 & -7 & -4 \\ 3 & -2 & 1 \\ 1 & 1 & \lambda \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
We can expand the determinant:
\[ = 3 \begin{vmatrix} -2 & 1 \\ 1 & \lambda \end{vmatrix} - (-7) \begin{vmatrix} 3 & 1 \\ 1 & \lambda \end{vmatrix} - 4 \begin{vmatrix} 3 & -2 \\ 1 & 1 \end{vmatrix} \]

Now calculate each of the 2x2 determinants:
- For \( \begin{vmatrix} -2 & 1 \\ 1 & \lambda \end{vmatrix} = -2\lambda - 1 \)
- For \( \begin{vmatrix} 3 & 1 \\ 1 & \lambda \end{vmatrix} = 3\lambda - 1 \)
- For \( \begin{vmatrix} 3 & -2 \\ 1 & 1 \end{vmatrix} = 3 \cdot 1 - (-2) \cdot 1 = 3 + 2 = 5 \)

Substitute these back into the determinant:
\[ = 3(-2\lambda - 1) + 7(3\lambda - 1) - 4(5) \]
\[ = -6\lambda - 3 + 21\lambda - 7 - 20 \]
\[ = (21\lambda - 6\lambda) + (-3 - 7 - 20) \]
\[ = 15\lambda - 30 \]

4. **Set the Determinant to Zero**:
Set the equation to zero:
\[ 15\lambda - 30 = 0 \]

5. **Solve for \( \lambda \)**:
\[ 15\lambda = 30 \]
\[ \lambda = \frac{30}{15} = 2 \]

### Final Answer:
Thus, the value of \( \lambda \) is \( \boxed{2} \).