Given the vectors $\vec{a}=\vec{i}+2\vec{j}-\vec{k}, \vec{b}=\vec{i}-\vec{j}+\vec{k}, \vec{c}=\lambda\vec{i}+\vec{k},\lambda\in\Bbb R$ in $V^{3}(0)$ find all scalars $\lambda$ for which the vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar.
$\alpha\vec{a}+\beta\vec{b}+\gamma\vec{c}=\vec{0}$
In general, for these vectors to be coplanar, does that mean $\alpha\neq 0, \beta\neq 0, \gamma\neq 0$ or is it sufficient that at least one of them is not 0?
Here's what I did:
$\alpha(\vec{i}+2\vec{j}-\vec{k})+\beta(\vec{i}-\vec{j}+\vec{k})+\gamma(\lambda\vec{i}+\vec{k})=\vec{0}$
$\Rightarrow$
$\alpha\vec{i}+\beta\vec{i}+\gamma\lambda\vec{i}=\vec{0}$,
$2\alpha\vec{j}-\beta{j}=\vec{0}$
$-\alpha\vec{k}+\beta\vec{k}+\gamma{k}=\vec{0}$
$\Rightarrow$
$\vec{i}(\alpha+\beta+\gamma\lambda)=0$
$\vec{j}(2\alpha-\beta)=0$
$\vec{k}(-\alpha+\beta+\gamma)=0$
$\Rightarrow$
$\alpha+\beta+\lambda\gamma=0$
$2\alpha-\beta=0$
$-\alpha+\beta+\gamma=0$
In the end I got $\beta=2\alpha, \gamma(\lambda-3)=0$
Now if I put $\gamma=0$ then that would mean $\alpha=0$ and that would mean $\beta=0$. So my vectors would be non-coplanar.
So I have to put $\lambda=3$.
Is that the final solution?