Answer :

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


2 + 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.


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