If `veca, vecb, vecc` are non coplanar vectors and `lamda` is a real number, then the vectors `veca+2vecb+3vecc, lamdavecb+4vec` and `(2lamda-1)vecc` are non coplanar for

Options

A. `mu in R`
B. `lamda = (1)/(2)`
C. `lamda =0`
D. no value of `lamda`

Explanation

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 can set up a determinant based on the coefficients of these vectors.

### Step-by-step Solution:

1. **Identify the Vectors and Their Coefficients:**
- The first vector is \(\vec{a} + 2\vec{b} + 3\vec{c}\) with coefficients:
- Coefficient of \(\vec{a}\) = 1
- Coefficient of \(\vec{b}\) = 2
- Coefficient of \(\vec{c}\) = 3
- The second vector is \(\lambda \vec{b} + 4\vec{c}\) with coefficients:
- Coefficient of \(\vec{a}\) = 0
- Coefficient of \(\vec{b}\) = \(\lambda\)
- Coefficient of \(\vec{c}\) = 4
- The third vector is \((2\lambda - 1)\vec{c}\) with coefficients:
- Coefficient of \(\vec{a}\) = 0
- Coefficient of \(\vec{b}\) = 0
- Coefficient of \(\vec{c}\) = \(2\lambda - 1\)

2. **Set Up the Determinant:**
To check for non-coplanarity, we need to compute the determinant of the matrix formed by these coefficients:
\[ \begin{vmatrix} 1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2\lambda - 1 \end{vmatrix} \]

3. **Calculate the Determinant:**
The determinant can be calculated as follows:
\[ \text{Determinant} = 1 \cdot (\lambda \cdot (2\lambda - 1) - 0) - 2 \cdot (0) + 3 \cdot (0) \]
This simplifies to:
\[ \lambda(2\lambda - 1) \]

4. **Set the Determinant to Zero:**
For the vectors to be non-coplanar, the determinant must not equal zero:
\[ \lambda(2\lambda - 1) = 0 \]

5. **Solve for \(\lambda\):**
Setting the equation to zero gives us:
- \(\lambda = 0\)
- \(2\lambda - 1 = 0 \Rightarrow \lambda = \frac{1}{2}\)

6. **Conclusion:**
The vectors are non-coplanar for \(\lambda = 0\) or \(\lambda = \frac{1}{2}\). The value of \(\mu\) does not affect the coplanarity condition since it does not appear in the determinant.

### Final Answer:
The vectors are non-coplanar for \(\lambda = 0\) or \(\lambda = \frac{1}{2}\) and \(\mu\) can take any real value.