We are given with two planes,

Also, Perpendicular distance of the plane from origin = 1

We need to find the equation of such plane.

We know,

Simplify the planes,

⇒ x + 3y – 6 = 0 …(i)

Also, for

⇒ 3x – y – 4z = 0 …(ii)

We know that,

The equation of a plane through the line of intersection of the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

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

Similarly, equation of a plane through the line of intersection of the planes x + 3y – 6 = 0 and 3x – y – 4z = 0 is

(x + 3y – 6) + λ(3x – y – 4z) = 0

⇒ x + 3y – 6 + 3λx – λy – 4λz = 0

⇒ x + 3λx + 3y – λy – 6 – 4λz = 0

⇒ (1 + 3λ)x + (3 – λ)y – 4λz – 6 = 0 …(iii)

Also, we know that

Perpendicular distance of a plane, ax + by + cz + d = 0 from the origin is P, such that

Similarly, perpendicular distance of a plane (iii), which is equal to 1 (according to the question) is

Squaring on both sides, we get

⇒ (1 + 3λ)² + (3 – λ)² + (-4λ)² = 36

⇒ (1)² + (3λ)² + 2(1)(3λ) + (3)² + (λ)² – 2(3)(λ) + 16λ² = 36

[∵, (a + b)² = a² + b² + 2ab and (a – b)² = a² + b² – 2ab]

⇒ 1 + 9λ² + 6λ + 9 + λ² – 6λ + 16λ² = 36

⇒ 9λ² + 16λ² + λ² + 6λ – 6λ = 36 – 1 – 9

⇒ 26λ² + 0 = 26

⇒ 26λ² = 26

⇒ λ² = 1

⇒ λ = ± 1

First, substitute λ = 1 in equation (iii) to find the equation of the plane.

(1 + 3λ)x + (3 – λ)y – 4λz – 6 = 0

⇒ (1 + 3(1))x – (3 – 1)y – 4(1)z – 6 = 0

⇒ 4x – 2y – 4z – 6 = 0

Now, substitute λ = -1 in equation (iii) to find the equation of the plane.

(1 + 3λ)x + (3 – λ)y – 4λz – 6 = 0

⇒ (1 + 3(-1))x + (3 – (-1))y – 4(-1)z – 6 = 0

⇒ (1 – 3)x + (3 + 1)y + 4z – 6 = 0

⇒ -2x + 4y + 4z – 6 = 0

Thus, equation of the required plane is 4x – 2y – 4z – 6 = 0 and -2x + 4y + 4z – 6 = 0.