Find $t\in \Bbb R$ such that $3$ vectors are non coplanar

Question

For which $t\in \Bbb R$ are vectors $\vec{a}=\vec{i}+2\vec{j}+3\vec{k}$ , $\vec{b}=\vec{i}+t\vec{j}+\vec{k}$ and $\vec{c}=\vec{i}+\vec{j}+\vec{k}$ non coplanar?

Vectors are non coplanar if and only if $\alpha\vec{a}+\beta\vec{b}+\gamma\vec{c}=0 \iff \alpha=\beta=\gamma=0$

So in this case $\alpha(\vec{i}+2\vec{j}+3\vec{k})+\beta(\vec{i}+t\vec{j}+\vec{k})+\gamma(\vec{i}+\vec{j}+\vec{k})=0 \iff \alpha=\beta=\gamma=0$

$\alpha\vec{i}+2\alpha\vec{j}+3\alpha{k}+\beta\vec{i}+\beta t\vec{j}+\beta\vec{k}+\gamma\vec{i}+\gamma\vec{j}+\gamma\vec{k}=0$

I know I should do this, but I don't understand why I'm doing it:

$(\alpha+\beta+\gamma)\vec{i}+(2\alpha+t\beta+\gamma)\vec{j}+(3\alpha+\beta+\gamma)\vec{k}=0$

$\alpha+\beta+\gamma=0$

$2\alpha+t\beta+\gamma=0$

$3\alpha+\beta+\gamma=0$

What are we trying to accomplish here? And how do I solve this system?

Answer 1

From the first equation and the third equation it is obtained that $\alpha$ is $0$. Then, we also obtain that $\beta=-\gamma$. So, the middle equation helps us to find out $t$.

$0+t\cdot \beta + \gamma=0$, $t \cdot \beta - \beta=0$, $\beta(t-1)=0$. So, $\beta=0$ or $t=1$.

Answer 2

For this system not to have trivial solution only (0,0,0), determinant of the matrix has to be 0

$$αtβγ+3αβγ+2αβγ-αβγ-2αβγ-3αtβγ=0 / :αβγ$$ $$-2t+2=0 $$ $$t=1$$