If \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors and \(\lambda\) is a real number, then the vectors \(\bar{a}+2 \bar{b}+3 \bar{c}, \lambda \bar{b}+4 \bar{c}\) and \((2 \lambda-1) \bar{c}\) are non-coplanar for
Step-by-step Solution:
Vectors are no coplanar if \[\] \(\left|\begin{array}{ccc}1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2 \lambda-1\end{array}\right| \neq 0\) \[\] \(\Rightarrow(2 \lambda-1)(\lambda) \neq 0 \Rightarrow \lambda \neq \frac{1}{2}\) and \(\lambda \neq 0 \quad\) Choice (C)\[\]