NCERT Solutions
Class 12 Maths
Three Dimensional Geometry

Ex.Misc.Q13
Find the equation of the plane passing through the point (−1, 3, 2) and
perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
The equation of the plane passing through the point (−1, 3, 2) is
a(x + 1) + b(y − 3) + c (z − 2) = 0 ......... 1
where, a, b, c are the direction ratios of normal to the plane.
It is known that two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are perpendicular, if a1a2 + b1b2 + c1c2 = 0
Plane 1 is perpendicular to the plane x + 2y + 3z = 5
So, a * 1 + b * 2 + c * 3 = 0
=> a + 2b + 3c = 0 ………..2
Also, plane 1 is perpendicular to the plane 3x + 3y + z = 0
So, a * 3 + b * 3 + c * 1 = 0
=> 3a + 3b + c = 0 ………..3
From equations 2 and 3, we get
=>
=> a-7=b8=c-3= k (say)
=> a = -7k, b = 8k, c = -3k
Substituting the values of a, b, and c in equation 1, we get
-7k(x + 1) + 8k(y - 3) – 3k(z - 2) = 0
=> -7(x + 1) + 8(y - 3) – 3(z - 2) = 0
=> -7x – 7 + 8y – 24 – 3z + 6 = 0
=> 7x – 8y + 3z + 25 = 0
This is the required equation of the plane.