Given $\mathbf{a} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix},$ $\mathbf{b} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix},$ and $\mathbf{c} = \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix},$ compute
\[(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} - \mathbf{a} \times (\mathbf{b} \times \mathbf{c}).\]
We have that
\begin{align*}
(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} &= \left( \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right) \times \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix} \\
&= \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix} \\
&= \begin{pmatrix} 6 \\ 3 \\ 0 \end{pmatrix}
\end{align*}and
\begin{align*}
\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) &= \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \left( \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix} \right) \\
&= \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \\
&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.
\end{align*}Hence,
\[(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} - \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \boxed{\begin{pmatrix} 6 \\ 3 \\ 0 \end{pmatrix}}.\]The main point of this exercise is to illustrate that in general,
\[(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} \neq \mathbf{a} \times (\mathbf{b} \times \mathbf{c}).\]In other words, the cross product is not associative.