The coordinates of the foot of the perpendicular drawn from a point P (-1,1,2) to the plane 2 x-3 y+z-11=0

Step-by-step solution

The foot of the perpendicular from point P(-1,1,2) to the plane 2x - 3y + z - 11 = 0 is determined by its intersection with the line through P parallel to the plane's normal vector n = (2,-3,1) . Using the parametric form x+1 2 = y-1 -3 = z-2 1 = k , the coordinates of a general point on this line are x = 2k - 1 , y = -3k + 1 , z = k + 2 . Substituting into the plane equation gives 2(2k - 1) - 3(-3k + 1) + (k + 2) - 11 = 0 , which simplifies to 14k - 14 = 0 , yielding k = 1 . The coordinates at k=1 are (1,-2,3) , m