A perpendicular is drawn from a point on the line x - 1 2 = y + 1 - 1 = z 1 to the plane x + y + z = 3 such that the foot of the perpendicular Q also lies on the plane x - y + z = 3 . Then the coordinates of Q are

Step-by-step solution

The given line is x - 1 2 = y + 1 - 1 = z 1 = λ ,   ( let ) ⇒ x - 1 2 = λ ,   y + 1 - 1 = λ ,   z 1 = λ ⇒ x = 2 λ + 1 ,   y = - λ - 1 ,   z = λ . Hence, any point on the line is 2 λ + 1 ,   - λ - 1 ,   λ . Thus, the point P on the line is 2 λ + 1 ,   - λ - 1 ,   λ . We know that the foot of perpendicular from a point x 1 ,   y 1 ,   z 1 on the plane a x + b y + c z + d = 0 is x - x 1 a