Answer :

Let P_{1} be the first plane and equation of P_{1} is

or x + 3y – 6 = 0

And P_{2} be the second plane and its equation is

or 3x – y – 4z = 0

As the equation of the plane passing through intersection of 2 planes P_{1} and P_{2} is given by

P_{1} + λP_{2} = 0

∴ x + 3y – 6 + λ(3x – y – 4z) = 0

⇒ x(1 + 3λ) + y(3 – λ) -4λz – 6 = 0

Given the distance of this plane from (0,0,0) is 1.

Length of perpendicular from a point (x_{1},y_{1},z_{1}) to a plane

ax + by + cz + d = 0 is given by –

p =

∴ length of perpendicular from (0, 0, 0) to plane

x(1 + 3λ) + y(3 – λ) -4λz – 6 = 0 is given by-

p =

As p = 1 (given)

Squaring both sides -

∴

⇒ 1 + 6λ + 9λ^{2} + 9 + λ^{2} – 6λ + 16λ^{2} = 36

⇒ 26λ^{2} + 10 = 36

⇒ 26λ^{2} = 26

⇒ λ^{2} = 1

⇒ λ = ±1

∴ equation of planes are given by –

x(1 + 3λ) + y(3 – λ) -4λz – 6 = 0

putting λ = 1

we have: 4x + 2y – 4z = 6

and putting λ = - 1

we have: -2x + 4y + 4z – 6 = 0

**OR**

Let P_{1} be the first plane and equation of P_{1} is

or x – y + 2z - 5 = 0

Normal vector to this plane is

And P_{2} be the second plane and its equation is

or 3x + y + z = 6

Normal vector to this plane is

As line is parallel to both the planes so it will be perpendicular to both normal vector and parallel to the cross product of

So we will first find

Expanding about the first row, we get –

⇒ =

As we have one point and got a line parallel to the line for which equation is asked. So we can give the equation easily.

Equation of the line is given by –

where is P.V of any random point on line. is the vector parallel to the line whose equation is given and is the P.V o any point on line.

∴ [given]

∴ equation of line is –

or

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