Start with $$m_1 \, \frac{d^2r_1}{dt^2} = F_{12} \qquad m_2 \, \frac{d^2r_2}{dt^2} = F_{21}$$ and derive the equation $$\frac{m_1m_2}{m_1 + m_2} \, \frac{d^2r}{dt^2} = F_{21} $$ where $r_1$ and $r_2$ are position vectors and $r = r_2 - r_1$.
Hint: Use Newton's Third law, which tells us that $F_{12} = -F_{21}$
I have tried several things such as collecting the terms of Newton's Third Law and solving the equation. I have also tried writing $F_{12} = -F_{21}$ in vector notation (i.e. $\langle a, b, c \rangle) $
As I have some information about classical mechanics this is called the two body problem. You have
$$\left\{ \matrix{ {{{d^2}{r_1}} \over {d{t^2}}} = {{{F_{21}}} \over {{m_1}}} \hfill \cr {{{d^2}{r_2}} \over {d{t^2}}} = {{{F_{21}}} \over {{m_2}}} \hfill \cr} \right.$$
Subtract the above equations considering ${F_{12}} = - {F_{21}}$ to get
$${{{d^2}{r_2}} \over {d{t^2}}} - {{{d^2}{r_1}} \over {d{t^2}}} = \left( {{1 \over {{m_2}}} + {1 \over {{m_1}}}} \right){F_{21}}$$
Hence by the definition of $r$ you have
$${{{d^2}} \over {d{t^2}}}\left( {{r_2} - {r_1}} \right) = {{{d^2}r} \over {d{t^2}}} = {{{m_1} + {m_2}} \over {{m_1}{m_2}}}{F_{21}}$$
and finally
$${{{m_1}{m_2}} \over {{m_1} + {m_2}}}{{{d^2}r} \over {d{t^2}}} = {F_{21}}$$