Provet the following, for four points whose position vectors are $\vec{\alpha}, \vec{\beta}, \vec{\gamma}, \vec{\delta}$ are co-planar Problem : Show that, the four points whose position vectors are $\vec{\alpha}, \vec{\beta}, \vec{\gamma}, \vec{\delta}$ are co-planar, iff $[\vec{\alpha}\vec{\beta}\vec{\gamma}]=[\vec{\beta}\vec{\gamma}\vec{\delta}]+[\vec{\gamma}\vec{\alpha}\vec{\delta}]+[\vec{\alpha}\vec{\beta}\vec{\delta}]$ 
Since $\vec{\alpha}, \vec{\beta}, \vec{\gamma}, \vec{\delta}$ are non-coplaner  can we write $x\vec{\alpha}+y\vec{\beta}+z\vec{\gamma}+t\vec{\delta}=\vec 0$  ?
 A: The four points whose position vectors are $\vec\alpha,\vec\beta,\vec\gamma,\vec\delta$ are co-planner$\iff$ the vectors  $\vec\beta-\vec\alpha,\vec\gamma-\vec\alpha,\vec\delta-\vec\alpha$ lie on a same plane $$\iff\big[(\vec\beta-\vec\alpha)\times(\vec\gamma-\vec\alpha)\big]\cdot(\vec\delta-\vec\alpha)=0$$$$\iff\big[(\vec\beta\times\vec \gamma)-(\vec\beta\times \vec \alpha)-(\vec\alpha\times\vec\gamma)+(\vec\alpha\times\vec\alpha)\big]\cdot(\vec\delta-\vec\alpha)=0$$$$\iff (\vec\beta\times \vec\gamma)\cdot\vec\delta-(\vec\beta\times \vec\gamma)\cdot\vec\alpha-(\vec\beta\times\vec\alpha)\cdot\vec\delta+(\vec\beta\times\vec\alpha)\cdot\vec\alpha-(\vec\alpha\times\vec\gamma)\cdot\vec\delta+(\vec\alpha\times\vec\gamma)\cdot\alpha=0$$$$\iff [\vec\beta\vec\gamma\vec\delta]-[\vec\alpha\vec\beta\vec\gamma]+[\vec\alpha\vec\beta\vec\delta]+0+[\vec\gamma\vec\alpha\vec\delta]+0=0.$$ Note that, in the third line we have used $\vec\alpha\times\vec\alpha=\vec0$.

$\textbf{Fact that I have used is }$$\vec a,\vec b,\vec c$ lie on a plane $\iff(\vec a \times\vec b)\cdot\vec c=0$. 
So let $\vec a,\vec b,\vec c$ lie on a plane, then we have some $(r,s,t)\not=(0,0,0)$ with $r\vec a+s\vec b+t\vec c=\vec 0$. Now $t\not=0\implies \vec c=-\frac{r}{t}\vec a-\frac{s}{t}\vec b\implies (\vec a \times\vec b)\cdot\vec c=-\frac{r}{t}(\vec a\times \vec b)\cdot \vec a-\frac{s}{t}(\vec a\times \vec b)\cdot b=0-0=0$. Next $s\not=0\implies\vec b=-\frac{r}{s}\vec a-\frac{t}{s}\vec c\implies (\vec a\times \vec c)\cdot \vec b=-\frac{r}{s}(\vec a\times \vec c)\cdot \vec b-\frac{t}{s}(\vec a\times \vec c)\cdot \vec b=0-0=0$, so that $\vec b\cdot (\vec a\times \vec c)=0\implies (\vec b\times \vec a)\cdot \vec c=0\implies (\vec a\times \vec b)\cdot \vec c=0$. Similarly for the case, $r\not=0$. So we are done only if part.
For the if part, consider $(\vec a\times \vec b)\cdot \vec c=0$. Now if $\vec a,\vec b$ are parallel then any plane containg $\vec a,\vec c$ must also contains $\vec c$, so we are done. For the other case when $\vec a,\vec b$ are not parallel then, $\{\vec a,\vec b,\vec a\times \vec b\}$ forms an orthogonal basis for $\Bbb R^3$, so we can write $\vec c=(\vec a\cdot\vec c)\vec a+(\vec b\cdot\vec c)\vec b+\big(\vec(a\times \vec b)\cdot \vec c\big)=(\vec a\cdot\vec c)\vec a+(\vec b\cdot\vec c)\vec b+0\implies\vec c$ is a linear combination of $\vec a,\vec b$. So we are done.  
