Answer
$-20$
Work Step by Step
To evaluate an nth-order determinant, where $n > 2$,
1. Select a row or column about which to expand.
2. Multiply each element $a_{ij}$ in the row or column by $(-1)^{i+j}$ times the determinant obtained by deleting the ith row and the $j\mathrm{t}\mathrm{h}$ column in the given array of numbers.
3. The value of the determinant is the sum of the products found in step 2.
-----
We select column 2, (all 2s ).
The first position, (i,j)=(1,2), 1+2=3,
defines the sequence of alternating signs starting with a minus.
$\left|\begin{array}{lll}
1 & 2 & 3\\
2 & 2 & -3\\
3 & 2 & 1
\end{array}\right|=$
$=-2\cdot\left|\begin{array}{ll}
2 & -3\\
3 & 1
\end{array}\right|+2\cdot\left|\begin{array}{ll}
1 & 3\\
3 & 1
\end{array}\right|-2\cdot\left|\begin{array}{ll}
1 & 3\\
2 & -3
\end{array}\right|$
$=2[-(2+9)+(1-9)-(-3-6)]$
$=2(-11-8+9)$
$=2(-10)$
$=-20$
