Three points are collinear if their determinant is equal to 0
So,
| 0 -10 1 |
| -3 -15 1 |
| 2 -6 1 |
We must put 1 at the third column since we don't have z in these points, and 1 is a neutral factor.
Applying sarrus
( 0 . -15 . 1 + -10 . 1 . 2 + 1 . -3 . -6) - (1 . -15 . 2 + 0 . 1 . -6 + -10 . -3 . 1)
(0 - 20 + 18) - (-30 + 0 + 30)
(-2) - (0)
-2
Our determinant is different than 0, so these points are not collinear.
