The determinant of the matrix \( \Delta \) is calculated as: \[ \Delta = \begin{vmatrix} 3 & 2 & 3 \\ 2 & 2 & 3 \\ 3 & 2 & 3 \end{vmatrix} = 3 \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} - 2 \begin{vmatrix} 2 & 3 \\ 3 & 3 \end{vmatrix} + 3 \begin{vmatrix} 2 & 2 \\ 3 & 2 \end{vmatrix} \] Now calculate each 2x2 determinant: \[ \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = 2 \cdot 3 - 3 \cdot 2 = 0 \] \[ \begin{vmatrix} 2 & 3 \\ 3 & 3 \end{vmatrix} = 2 \cdot 3 - 3 \cdot 3 = 6 - 9 = -3 \] \[ \begin{vmatrix} 2 & 2 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 - 2 \cdot 3 = 4 - 6 = -2 \] Substitute these values back into the original determinant expression: \[ \Delta = 3(0) - 2(-3) + 3(-2) = 0 + 6 - 6 = 0 \] Thus, \[ |\Delta| = 0 \] Final Answer: \[ \boxed{0} \]