Q27 of 29 Page 1

Find the distance between the point P (6, 5, 9) and the plane determined by the points A (3, -1, 2), B (5, 2, 4) and C (-1, -1, 6).

A, B and C are 3 points on the plane.


Normal to the plane, N = (A – B) × (A – C)





So, the Cartesian equation of the plane is of the form


⇒ 12x - 16y + 12z = d


Since (3, -1, 2) is a point on the plane,


⇒ 36 + 16 + 24 = d


⇒ d = 76


Equation of plane is


⇒ 3x – 4y + 3z – 19 = 0


Distance of point P (6, 5, 9) from the plane is




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