Question (Here, $\wedge$ is the cross product for vectors. You might have also seen it as $\times$):
Show that the equation of the plane containing three non-collinear points with position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is $\mathbf{r} \cdot (\mathbf{a} \wedge \mathbf{b} + \mathbf{b} \wedge \mathbf{c} + \mathbf{c} \wedge \mathbf{a}) = [\mathbf{a},\mathbf{b},\mathbf{c}]$
Right now, I have no idea on how to begin the question. In a previous question, we proved that $A,B,C$ are collinear if and only if $\mathbf{a} \wedge \mathbf{b} + \mathbf{b} \wedge \mathbf{c} + \mathbf{c} \wedge \mathbf{a} = \mathbf{0}$, and I know that I can distribute $\mathbf{r}$ over $(\mathbf{a} \wedge \mathbf{b} + \mathbf{b} \wedge \mathbf{c} + \mathbf{c} \wedge \mathbf{a})$. Any hints, directions, or answers would be greatly appreciated.
Note: This is for a first-year Geometry course in University.
This follows directly from the equation of the plane in the form $$\underline{r}\cdot\underline{n}=\underline{a}\cdot\underline{n}$$
Where $\underline{n}$ is the vector normal to the plane.
That is, $$\underline{n}=(\underline{b}-\underline{a})\times(\underline{c}-\underline{a})$$ And this is equivalent to $$\underline{a}\times\underline{b}+\underline{b}\times\underline{c}+\underline{c}\times\underline{a}$$
Thus, $$\underline{a}\cdot\underline{n}=\underline{a}\cdot\underline{b}\times\underline{c}$$
And hence the result.