If $A,B,C$ are collinear, prove $\vec{A}\times \vec{B} + \vec{B}\times \vec{C} + \vec{C}\times \vec{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$ Prove that if $A, B$ and $C$ are collinear, then
$\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$.
So far, I know that $\vec{AB}, \vec{BC}, \vec{AC}$ are scalar multiples and their cross product is zero. I'm not sure how to apply this to prove $\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A}$ though.

Setting up some vector identities:
$$ 
\overrightarrow{AB} = \overrightarrow{B} -\overrightarrow{A} \\
\overrightarrow{A} \times \overrightarrow{B} = - \overrightarrow{B} \times \overrightarrow{A} \\
\overrightarrow{A} \times \overrightarrow{A} = \overrightarrow{0}
$$
Proof:
$$
\begin{align}
&\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A} \\
&= \overrightarrow{A} \times \overrightarrow{B} - \overrightarrow{A} \times \overrightarrow{C} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{A} \times (\overrightarrow{B} - \overrightarrow{C}) + \overrightarrow{B} \times \overrightarrow{C} \\
&= (\overrightarrow{B} - \overrightarrow{AB}) \times \overrightarrow{CB} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{B} \times \overrightarrow{CB} - \overrightarrow{AB} \times \overrightarrow{CB} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{B} \times (\overrightarrow{CB} + \overrightarrow{C}) - \overrightarrow{0} \\
&= \overrightarrow{B} \times (\overrightarrow{B} - \overrightarrow{C} + \overrightarrow{C}) \\
&= \overrightarrow{0}
\end{align}
$$
