Condition For Line Lying In A Plane

The test involves two conditions: a line lies in a plane only if its direction vector is perpendicular to the plane's normal (so b \cdot n = 0) AND a point on the line satisfies the plane equation. Both must hold, a nuance regularly examined in JEE Advanced. The direction is b = (2, -1, 4) and the normal is n = (1, 2, -1). Their dot product is 2(1) + (-1)(2) + 4(-1) = 2 - 2 - 4 = -4, which is nonzero, so the direction is not perpendicular to the normal. Checking the point (1, 2, 3): 1 + 2(2) - 3 = 1 + 4 - 3 = 2, which does satisfy the plane, so the line passes through this point of the plane. But because b \cdot n \neq 0, the line is oblique to the plane and crosses it; hence it meets the plane at exactly one point rather than lying in it. The yes option fails because b \cdot n \neq 0. The parallel-but-outside option fails because parallelism would need b \cdot n = 0 with the point off the plane, whereas the point is on the plane. The perpendicular option would require b parallel to n, which is false. This applies the line-in-plane criterion theorem. Plausibility check: a line through a point of the plane whose direction has a nonzero component along the normal must pierce the plane at that single point, exactly one intersection.