The equation of the plane passing through the intersection of the planes 2x + y + 2z = 9, 4x – 5y – 4z = 1 and the point (3, 2, 1) is

Concept:

If a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 represents two different planes, then equation of plane passing through the intersection of these planes is given by:

(a₁x + b₁y + c₁z + d₁) + λ × (a₂x + b₂y + c₂z + d₂) = 0.

Calculation:

Given: Two planes 2x + y + 2z – 9 = 0 and 4x – 5y – 4z – 1 = 0.

As we know that, if a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 represents two different planes, then equation of plane passing through the intersection of these planes is given by:

(a₁x + b₁y + c₁z + d₁) + λ × (a₂x + b₂y + c₂z + d₂) = 0.

So, the plane passing through the intersection of two given planes is:

⇒ (2x + y + 2z – 9) + λ × (4x – 5y – 4z – 1) = 0     ...1)

∵ it is given that the plane passing through the intersection of two given plane also passes through the point (3, 2, 1).

⇒ The point (3, 2, 1) will satisfy the equation (1)

⇒ (6 + 2 + 2 - 9) + λ × (12 – 10 – 4 - 1) = 0 ⇒ λ = 1/3.

So, by substituting the value of λ in equation (1), we get