UPSC NDA Math Model Paper 6

Given: Equation of planes: x + y + z = 6 and 2x + 3y + 4z + 5 = 0
The given equations can be re-written as: x + y + z - 6 = 0 and 2x + 3y + 4z + 5 = 0
As we know that, the equation of a plane passing through the line of intersection of two planes ax+by+cz+d=0 and ax+by+cz+d=0 is given by: (a1x+b1y+c1z+d1) +λ(a2x+b2y+c2z+d2)=0 where λ∈R
⇒(x+y+z−6)+λ(2x+3y+4z+5)=0 .........(1)
∵ It is given that plane passing through the line line of intersection of the plane x + y + z = 6 and 2x + 3y + 4z + 5 = 0 also passes through the point (1, 1, 1)
i.e The point (1, 1, 1) will satisfy the equation (1)
⇒ (1 + 1 + 1 - 6) + λ (2 + 3 + 4 + 5) = 0
⇒ - 3 + 14λ = 0
⇒ λ = 3/14
Now by substituting λ = 3/14 in equation (1) we get
⇒(x+y+z−6)+

3

14

.(2x+3y +4z+5)=0
⇒20x+23y+26z−69=0
Hence, the equation of the required plane is: 20x+23y+26z−69=0