This is called

expansion by minors. (You can use the Leibniz formula to make a nice compact formula, but I've found that to be difficult to put into use unless I'm writing a computer program to do determinants.)

Uck. This is going to be awful to type!

$\displaystyle \left | \begin{matrix} 2 & 1 & 2 & 1 \\ 3& 0 & 1 & 1 \\ -1 & 2 & -2 & 1 \\ -3 & 2 & 3 & 1 \end{matrix} \right | = 2 \left | \begin{matrix} 0 & 1 & 1 \\ 2 & -2 & 1 \\ 2 & 3 & 1 \end{matrix} \right | - 1 \left | \begin{matrix} 3 & 1 & 1 \\ -1 & -2 & 1 \\ -3 & 3 & 1 \end{matrix} \right |$ $\displaystyle + 2 \left | \begin{matrix} 3 & 0 & 1 \\ -1 & 2 & 1 \\ -3 & 2 & 1 \end{matrix} \right | - 1 \left | \begin{matrix} 3 & 0 & 1 \\ -1 & 2 & -2 \\ -3 & 2 & 3 \end{matrix} \right |$

I presume you know how to do 3x3 determinants. If not, do an expansion by minors on each of the 4 3x3 determinants listed here.

-Dan