Equation of a plane in vector form I have a basic doubt. If we are given 3 points in space then the equation of the plane passing through them is given by:
$$ \Bigg[\vec{r} - \vec{a}, \vec{a} - \vec{b}, \vec{a} - \vec{c}\Bigg] = 0$$
where $ \vec{r} $ is the position vector of the variable point and $\vec{a}, \vec{b}, \vec{c}$ are the fixed points. This equation is true because the box product of coplanar vectors is $0$. However, if we see then even this:
$$ \Bigg[\vec{r} - \vec{a}, \vec{r} - \vec{b}, \vec{r} - \vec{c}\Bigg] = 0$$
satisfies the condition - the box product of 3 coplanar vectors is zero. However, if we right this is Cartesian form, then we don't get a linear equation and that wouldn't represent a plane, would it?
 A: As commented earlier, box product is rusty for me, but looks to me that the expressions are the same. 
$$\Big[\vec{r} - \vec{a}, \vec{r} - \vec{b}, \vec{r} - \vec{c}\Big] = 0$$
is equivalent to 
$$(\vec{r} - \vec{a}) \times (\vec{r} - \vec{b})\cdot(\vec{r} - \vec{c}) = 0$$
$$(\vec{r}\times \vec{r} - \vec{r} \times \vec{b} - \vec{a}\times\vec{r} +\vec{a}\times\vec{b} )\cdot(\vec{r} - \vec{c}) = 0$$
$$(- \vec{r} \times \vec{b} - \vec{a}\times\vec{r} +\vec{a}\times\vec{b} )\cdot(\vec{r} - \vec{c}) = 0$$
Noting $\vec{r} \times \vec{r}= 0$ and expanding, 
$$[\vec{r}, \vec{b}, \vec{c}]+[\vec{a}, \vec{r}, \vec{c}]+[\vec{a}, \vec{b}, \vec{r}]-[\vec{a}, \vec{b}, \vec{c}] = 0$$
So this should not lead to a non-linear equation!  Further, 
I leave testing if the first expression expands to the same to you...
A: A plane in $\mathbb R^3$ is a codimension $1$-linear subspace of $\mathbb R^3$, whose general form in cartesian coordinates is given by 
$$ax+by+cz+d=0,$$
with $a,b,c,d\in\mathbb R$. The second box-plot equation you wrote is not linear in the coordinates $(x,y,z)$ of $r$, so it does not represent a plane.
EDIT: The above statement in the case of the box equation is not true, as the quadratic terms in the components of $r$ vanish, as shown in Macavity's answer.
The following caveat remains valid.
In general, note that the points
$$a=(1,0,0),$$
$$b=(0,1,0),$$
$$c=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0)$$
are co-planar and satisfy the equation
$$x^2+y^2-1=0;$$
this latter does not describe a plane in $\mathbb R^3$, however. In other words, given an equation in the  coordinates $(x,y,z)$ which is satisfied by co-planar triples $a,b,c$ does not imply that the given equation describes a plane in $\mathbb R^3$.
A: Let $V$ be the volume of the tetrahedron with vertices $\vec{r}$, $\vec{a}$, $\vec{b}$, $\vec{c}$. This tetrahedron has 6 sides, namely the vectors $\vec{r} - \vec{a}$, $\vec{r} - \vec{b}$, $\vec{r} - \vec{c}$, $\vec{b} - \vec{a}$, $\vec{c} - \vec{a}$, $\vec{b} - \vec{c}$. The "box product" of any three of these 6 vectors gives $6V$ (which is actually the volume of a parallelipiped). You get the same result no matter which three of the six you use.
So, in particular, using the two box products you mentioned, we have
$$
\Big[\vec{r} - \vec{a}, \vec{r} - \vec{b}, \vec{r} - \vec{c}\Big] = 
\Big[\vec{r} - \vec{a}, \vec{a} - \vec{b}, \vec{a} - \vec{c}\Big]
$$
Macavity's answer shows algebraically that this is true. My only point is that it is (fairly) obvious geometrically, too, if you understand how box products relate to volumes.
