Collinearity when $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$ 
Let $\mathbf{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
  Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if
  $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}.$

Since the cross product of two vectors gives an area, and for two vectors to give an area of $0$ they need to be on the same line (or they can be a point, but I'm assuming both are not $\mathbf 0$). However, in this problem, each of the cross products need not necessarily be $0$ since it's their sum that is $0$, and now I'm not sure what to do. 
 A: Using that $\,\color{blue}{\mathbf{b} \times \mathbf{c} = -\,\mathbf{c} \times \mathbf{b}}\,$ and $\,\color{red}{\mathbf{a} \times \mathbf{a} = 0}\,$:
$$
\begin{align}
\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0} \;\;&\iff\;\; \mathbf{a} \times \mathbf{b} \color{blue}{- \mathbf{c} \times \mathbf{b}} + \mathbf{c} \times \mathbf{a} \color{red}{-  \mathbf{a} \times \mathbf{a}} = \mathbf{0} \\
 & \iff\;\; (\mathbf{a} - \mathbf{c}) \times \mathbf{b} + (\mathbf{c} - \mathbf{a}) \times \mathbf{a} = 0 \\
 & \iff\;\; (\mathbf{a} - \mathbf{c}) \times \mathbf{b} - (\mathbf{a} - \mathbf{c}) \times \mathbf{a} = 0 \\
 & \iff\;\; (\mathbf{a} - \mathbf{c}) \times (\mathbf{b} - \mathbf{a}) = 0
\end{align}
$$
The latter equality is equivalent to $\,\mathbf{a} - \mathbf{c}\,$ and $\, \mathbf{b} - \mathbf{a}\,$ being collinear.
A: Let $A$, $B$ and $C$ be the corresponding points. Do you know that they are colinear iff the vectors $b-a$ and $c-a$ are colinear? And as you said, $b-a$ and $c-a$ are colinear iff their cross product is $0$.
Edit
Let $O=(0,0,0)$ be the origin of $\mathbb{R}^3$. Then we have $\vec{OA}=a$, $\vec{OB}=b$ and $\vec{OC}=c$.
$A,B$ and $C$ are aligned iff $\vec{AB}$ and $\vec{AC}$ are parallel, and you can write $\vec{AC}=\lambda\vec{AB}$ for some $\lambda\in\mathbb{R}$.
Here's an example where $\lambda>0$

Here's an example where $\lambda<0$

If you agree that $A,B$ and $C$ are aligned iff $\vec{AB}$ and $\vec{AC}$ are parallel, notice that:
\begin{align*}
\vec{AB}&=\vec{AO}+\vec{OB}\\
&=\vec{OB}-\vec{OA}\\
&=b-a.
\end{align*}
Similarly, $\vec{AC}=c-a$.
Hence $A$, $B$ and $C$ are aligned iff $b-a$ and $c-a$ are parallel. 
The knowledge that these two vectors are parallel is enough. Actually, $c-b=(c-a)-(b-a)$. Since $\vec{AC}=\lambda\vec{AB}$, in other words, $c-a=\lambda(b-a)$, we have $c-b=(\lambda-1)(b-a)$ and thus $c-a$ is indeed parallel to both $b-a$ and $c-a$. Mentioning $c-b$ as well is not wrong, but redundant.
A: If the vectors are collinear, they are of the form
$$
\mathbf{a}=\mathbf{u}+\alpha\mathbf{v},
\quad
\mathbf{b}=\mathbf{u}+\beta\mathbf{v},
\quad
\mathbf{c}=\mathbf{u}+\gamma\mathbf{v}
$$
Then
$$
\mathbf{a}\times\mathbf{b}=
\mathbf{u}\times\mathbf{u}+\alpha\mathbf{v}\times\mathbf{u}+
\beta\mathbf{u}\times\mathbf{v}+\alpha\gamma\mathbf{v}\times\mathbf{v}=
(\beta-\alpha)\mathbf{u}\times\mathbf{v}
$$
Similarly $\mathbf{b}\times\mathbf{c}=(\gamma-\beta)\mathbf{u}\times\mathbf{v}$ and $\mathbf{c}\times\mathbf{a}=(\alpha-\gamma)\mathbf{u}\times\mathbf{v}$, so easily $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$.
Conversely, suppose $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$ and set $\mathbf{b}-\mathbf{a}=\mathbf{x}$. Without loss of generality, we can assume $\mathbf{a}\ne\mathbf{b}$ (or there would be just one or two vectors, which are obviously collinear). Then
\begin{align}
\mathbf{0}
&=\mathbf{a}\times(\mathbf{a}+\mathbf{x})
 +(\mathbf{a}+\mathbf{x})\times\mathbf{c}
 +\mathbf{c}\times\mathbf{a} \\[4px]
&=\mathbf{a}\times\mathbf{x}
 +\mathbf{a}\times\mathbf{c}
 +\mathbf{x}\times\mathbf{c}
 +\mathbf{c}\times\mathbf{a} \\[4px]
&=\mathbf{x}\times(\mathbf{c}-\mathbf{a})
\end{align}
which implies $\mathbf{c}-\mathbf{a}=\delta\mathbf{x}$. Thus
$$
\mathbf{a}=\mathbf{a}+0\mathbf{x},
\quad
\mathbf{b}=\mathbf{a}+1\mathbf{x},
\quad
\mathbf{c}=\mathbf{a}+\delta\mathbf{x}
$$
