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. 

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
$$
\mathbf{a}=\mathbf{a}+0\mathbf{x},
\quad
\mathbf{b}=\mathbf{a}+1\mathbf{x},
\quad
\mathbf{c}=\mathbf{a}+\delta\mathbf{x}
$$
