1. **Stating the problem:** Calculate the determinant of the matrix $$A = \begin{pmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{pmatrix}$$ using the diagonal (Sarrus) method. 2. **Formula and rules:** For a 3x3 matrix $$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix},$$ the determinant is $$|A| = aei + bfg + cdh - ceg - bdi - afh.$$ This means we multiply the diagonals from top-left to bottom-right and sum them, then subtract the sum of the diagonals from top-right to bottom-left. 3. **Apply the formula:** Identify elements: $$a=6, b=-3, c=2, d=2, e=-1, f=2, g=-10, h=5, i=2.$$ Calculate the positive diagonal products: $$6 \times (-1) \times 2 = -12,$$ $$(-3) \times 2 \times (-10) = 60,$$ $$2 \times 2 \times 5 = 20.$$ Sum these: $$-12 + 60 + 20 = 68.$$ Calculate the negative diagonal products: $$2 \times (-1) \times (-10) = 20,$$ $$6 \times 2 \times 5 = 60,$$ $$(-3) \times 2 \times 2 = -12.$$ Sum these: $$20 + 60 + (-12) = 68.$$ 4. **Calculate determinant:** $$|A| = 68 - 68 = 0.$$ 5. **Explanation:** The determinant is zero, which means the matrix is singular and does not have an inverse. **Final answer:** $$\boxed{0}$$