NCERT Solutions
Class 12 Maths
Three Dimensional Geometry

Ex.11.3 Q.6
Find the equation of the planes that passes through three points.
(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)
(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)
(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3).
Now, 1 1 -1
6 4 -5 = (12 - 10) – (18 - 20) – (-12 + 16) = 2 + 2 – 4 = 0
-4 -2 3
Since A, B, C are collinear points, there will be infinite number of planes passing through the given points.
(b) The given points are A (1, 1, 0), B (1, 2, 1), and C (−2, 2, −1).
Now, 1 1 0
1 2 1 = 1(-2 - 2) – 1(-1 + 2) + 0 = -4 – 1 = -5 ≠ 0
-2 2 -1
Therefore, a plane will pass through the points A, B, and C.
It is known that the equation of the plane through the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z2) is
x – x1 y – y1 z – z1
x2 – x1 y2 – y1 z2 – z1 = 0
x3 – x1 y3 – y1 z3 – z1
=> x - 1 y - 1 z
0 1 1 = 0
-3 1 -1
=> -2(x - 1) – 3(y – 1) + 3z = 0
=> -2x – 3y + 3z + 2 + 3 = 0
=> -2x – 3y + 3z = 5
=> 2x + 3y – 3z = 5
This is the Cartesian equation of the required plane.