Let P be the foot of the perpendicular drawn from the point Q(0, -1, 1) to the plane x + y + z = 6 . The perpendicular distance of the point P from the line x 2 = y-1 -1 = z-2 2 is equal to:

Step-by-step solution

First, we find the coordinates of P , the foot of the perpendicular from Q(0, -1, 1) to the plane x + y + z = 6 . The equation of the line passing through Q and perpendicular to the plane is: x - 0 1 = y + 1 1 = z - 1 1 = t A general point on this line is (t, t-1, t+1) . Since P lies on the plane, we substitute these coordinates into the plane equation: t + (t-1) + (t+1) = 6 3t = 6 t = 2 Thus, the coordinates of P are (2, 1, 3) . Next, we find the perpendicular distance of P(2, 1, 3) from the line L: x 2 = y-1 -1 =