Extend it like this:\(\begin{bmatrix} 1 && 2 &&-2 &&1&&2\\ 3 && 0 && -1&&3&&0\\ 5 && 3 && 2&&5&&3 \end{bmatrix}\)
Multiply the numbers if they can be linked top-to-bottom diagonally with 3 numbers:
Example: 1, 0, and 2 is in the same diagonal and they are 3 numbers.
Then add up the products obtained from multiplying the number of top-left to bottom-right. Call that 'answer 1'
(1)(0)(2) + (2)(-1)(5) + (-2)(3)(3) = 0 - 10 - 18 = -28 -----(1)
Then add up the products obtained from multiplying the number of top-right to bottom-left. Call that 'answer 2'
(-2)(0)(5) + (1)(-1)(3) + (2)(3)(2) = -10 - 3 + 12 = -1 -----(2)
Add up answer 1 and answer 2: -28 - 1 = -29.
Therefore det(C) = -29. Best method ever , I don't have a proof of it though but I got through the determinant questions correctly when I use this method.