IF `bara,barb,barc` are non-coplanar vectors and `lamda` is a real number then the vectors `bara+2barb+3barc,lamdabarb+4barc and(2lamda-1)barc` are non coplanar for

Options

A. all values of `lamda`
B. all except one value of `lamda`
C. all except two values of `lamda`
D. no value of `lamda`

Answer

To determine the values of \(\lambda\) for which the vectors \(\vec{a} + 2\vec{b} + 3\vec{c}\), \(\lambda \vec{b} + 4\vec{c}\), and \((2\lambda - 1)\vec{c}\) are non-coplanar, we need to check the condition for non-coplanarity, which is that the determinant of the coefficients of these vectors should be non-zero.

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The vectors can be expressed as:
- First vector: \(\vec{v_1} = \vec{a} + 2\vec{b} + 3\vec{c}\) has coefficients \( (1, 2, 3) \).
- Second vector: \(\vec{v_2} = \lambda \vec{b} + 4\vec{c}\) has coefficients \( (0, \lambda, 4) \).
- Third vector: \(\vec{v_3} = (2\lambda - 1)\vec{c}\) has coefficients \( (0, 0, 2\lambda - 1) \).

2. **Set Up the Determinant**:
The determinant of the matrix formed by these coefficients is given by:
\[ D = \begin{vmatrix} 1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2\lambda - 1 \end{vmatrix} \]

3. **Calculate the Determinant**:
Since the first column has zeros in the second and third rows, we can expand the determinant along the first column:
\[ D = 1 \cdot \begin{vmatrix} \lambda & 4 \\ 0 & 2\lambda - 1 \end{vmatrix} \]
The determinant of the 2x2 matrix is:
\[ D = 1 \cdot (\lambda(2\lambda - 1) - 0) = \lambda(2\lambda - 1) \]

4. **Set the Determinant Non-Zero**:
For the vectors to be non-coplanar, we require:
\[ D \neq 0 \implies \lambda(2\lambda - 1) \neq 0 \]
This implies:
- \(\lambda \neq 0\)
- \(2\lambda - 1 \neq 0 \implies \lambda \neq \frac{1}{2}\)

5. **Conclusion**:
The vectors \(\vec{a} + 2\vec{b} + 3\vec{c}\), \(\lambda \vec{b} + 4\vec{c}\), and \((2\lambda - 1)\vec{c}\) are non-coplanar for all values of \(\lambda\) except \(\lambda = 0\) and \(\lambda = \frac{1}{2}\).

### Final Answer:
The vectors are non-coplanar for:
\[ \lambda \in \mathbb{R} \setminus \{0, \frac{1}{2}\} \]