Intersection Of Line And Plane

The standard technique is parametrization: write the line in terms of a parameter t and substitute into the plane equation to solve for t, then back-substitute. This substitution method is the reliable JEE Advanced way to locate a line-plane intersection. Setting (x-1)/2 = (y+1)/3 = (z-2)/1 = t gives x = 1 + 2t, y = -1 + 3t, z = 2 + t. Plug into x + 2y + 3z = 14: (1 + 2t) + 2(-1 + 3t) + 3(2 + t) = 1 + 2t - 2 + 6t + 6 + 3t = 5 + 11t. Setting 5 + 11t = 14 gives 11t = 9, so t = 9/11. Back-substituting: x = 1 + 18/11 = 29/11, y = -1 + 27/11 = 16/11, z = 2 + 9/11 = 31/11. Thus the intersection point is (29/11, 16/11, 31/11). Option (3, 2, 3) corresponds to t = 1 but fails the plane. Option (5, 5, 4) uses t = 2 incorrectly. Option (2, 1, 3) is off the line entirely. This applies the parametric-substitution intersection method. Plausibility check: 29/11 + 2(16/11) + 3(31/11) = (29 + 32 + 93)/11 = 154/11 = 14, so the point satisfies the plane, and it lies on the line at t = 9/11 as required.