Collinearity Of Three Points

The criterion is that three points are collinear when the vectors joining them are parallel, that is one is a scalar multiple of the other, equivalently their cross product is the zero vector. This parallel-vector test is a clean JEE Advanced standard. Take A = (1, 2, 3), B = (2, 4, 6), C = (4, 8, 12). Then AB = B - A = (1, 2, 3) and AC = C - A = (3, 6, 9) = 3(1, 2, 3) = 3 \cdot AB. Since AC is a scalar multiple of AB, the vectors are parallel and share point A, so the three points lie on one line. Equivalently AB \times AC = (1,2,3) \times (3,6,9) = (2\cdot9 - 3\cdot6,\ 3\cdot3 - 1\cdot9,\ 1\cdot6 - 2\cdot3) = (0, 0, 0). The triangle option fails because zero cross product means zero area. The coplanar-not-collinear option is wrong since proportionality is stronger than coplanarity. No two points coincide, as all coordinates differ. This applies the collinearity-via-parallel-vectors theorem. Plausibility check: a vanishing cross product confirms zero enclosed area, the hallmark of collinear points.