Question
The equation of the plane containing the line 2x-5y +z = 3, x +y+4z = 5 and parallel to the plane x +3y +6z =1 is
-
2x + 6y + 12z = 13
-
x+3y+6z = -7
-
x+3y +6z = 7
-
2x+ 6y+12z = - 13
Solution
C.
x+3y +6z = 7
Let equation of plane containing the lines 2x- 5y +z = 3 and x+y+4z = 5 be
(2x-5y+z-3) + λ(x+y+4z-5) = 0
⇒ (2+λ)x + (λ-5)y + (4λ + 1)z -3 -5λ =0... (i)
This plane is parallel to the plane x +3y +6z = 1
On taking first two equalities, we get
6λ -30 = 3 + 12λ
-6λ = 33
λ = - 11/2
So, the equation of required plane is
