If $\vec{a},\vec{b},\vec{c}$ are non-coplanar vectors and $\lambda$ is a real number, then 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

Three vectors ${\vec{v}}_1,{\vec{v}}_2,{\vec{v}}_3$ are non-coplanar if their scalar triple product is non-zero. Let $[\vec{a},\vec{b},\vec{c}]\ne 0$.The triple product of the given vectors is:$\left|\begin{array}{ccc}1& 2& 3\\ 0& \lambda & 4\\ 0& 0& 2\lambda -1\end{array}\right|[\vec{a},\vec{b},\vec{c}]$.
The determinant value is $1(\lambda )(2\lambda -1)=\lambda (2\lambda -1)$.
For non-coplanarity, this determinant must not be zero:
$\lambda (2\lambda -1)\ne 0\Rightarrow \lambda \ne 0$ and $\lambda \ne 1/2$.
Thus, the vectors are non-coplanar for all real values of $\lambda$ except $0$ and $1/2$.