Foot Of Perpendicular From Point To Line

The approach parametrizes the line as a general point and imposes orthogonality of the connecting vector with the line's direction. This perpendicularity condition is the dependable JEE Advanced route to the foot of perpendicular. Let the line point be (t, 2t, 2t) from x/1 = y/2 = z/2 = t, with direction d = (1, 2, 2). The vector from the external point (1, 2, 3) to this foot is (t - 1, 2t - 2, 2t - 3). For perpendicularity, this must satisfy (t - 1)(1) + (2t - 2)(2) + (2t - 3)(2) = 0, giving t - 1 + 4t - 4 + 4t - 6 = 9t - 11 = 0, so t = 11/9. The foot is therefore (11/9, 22/9, 22/9). Option (2, 4, 4) corresponds to t = 2, too far along. Option (1, 1, 1) is not on the line. Option (0, 0, 0) is the origin, not the foot. This applies the perpendicular-foot orthogonality theorem. Plausibility check: the connecting vector at t = 11/9 is (2/9, 4/9, -5/9), and its dot with the direction (1, 2, 2) is 2/9 + 8/9 - 10/9 = 0, confirming perpendicularity, while the foot clearly lies on the line at parameter t = 11/9.