Answer
$-24$
Work Step by Step
$\begin{vmatrix}3&3&-3\\3&4&-4\\2&-3&-5\end{vmatrix}=\begin{vmatrix}3&3&-3\\0&1&-1\\0&-5&-3\end{vmatrix}=\begin{vmatrix}3&3&-3\\0&1&-1\\0&0&-8\end{vmatrix}=3\cdot 1\cdot (-8)=-24$
Note that we only added constant multiples of some rows to others, so the determinant of the echelon matrix equals the determinant of the original matrix.