Question

The equation of the planes passing through the line of intersection of the planes x + 3y + 6 = 0 = 3x − y − 4z which is at a distance of 1 unit from the origin are 2x + y − 2z + d₁ = 0 and x − 2y − 2z + d₂ = 0. Then − d₁d₂ is equal to _______.

Answer 1

The required plane is of the form

The distance of this plane provided in Eq. (1) from the origin = 1. This gives 

Therefore, the required planes are obtained by substituting λ = ±1 in Eq. (1) and they are 2x + y − 2z + 3 = 0 and x − 2y − 2z − 3 = 0. Therefore

- d₁d₂ = - 3(-3) = 9