Find the equation of plane passing through the line of intersection of planes 4x - 3y + 5z = 6 and x – y + z = 8 and is perpendicular to plane x – 4y – 2z = 9?

CONCEPT:

If a1 x + b1 y + c1 z + d = 0 and a2 x + b2 y + c2 z + d = 0 represents two different planes, then equation of a plane passing through the intersection of these planes is given by: (a1 x + b1 y + c1 z + d) + λ (a2 x + b2 y + c2 z + d) = 0, where λ is a scalar.

If two lines are perpendicular then a₁ ⋅ a₂ + b₁ ⋅ b₂ + c₁ ⋅ c₂ = 0

CALCULATION:

Here, we have to find the equation of plane passing through the line of intersection of planes 4x - 3y + 5z = 6 and x – y + z = 8 and is perpendicular to plane x – 4y – 2z = 9

As we know that, equation of a plane passing through the intersection of these planes is given by: (a1 x + b1 y + c1 z + d) + λ (a2 x + b2 y + c2 z + d) = 0, where λ is a scalar

⇒ (4x - 3y + 5z - 6) + λ (x - y + z - 8) = 0

⇒ (4 + λ)x - (λ + 3)y + (5 + λ)z - 6 - 8λ = 0---------(1)

The direction ratios of the plane represented by (1) are: (4 + λ), (- λ - 3), (5 + λ)

The direction ratios of the plane x – 4y – 2z = 9 are: 1, - 4, - 2

∵ plane represented by (1) and is perpendicular to plane x – 4y – 2z = 9

⇒ 4 + λ + 4λ + 12 - 10 - 2λ = 0

⇒ 3λ + 6 = 0

⇒ λ = - 2

By substituting λ = - 2 in equation (1), we get

⇒ 2x - y + 3z + 10 = 0

Hence, option A is the correct answer