Find the equation of the plane passing through the line of intersection of the planes and whose perpendicular distance from origin is unity.

OR

Find the vector equation of the line passing through the point (1, 2, 3) and parallel to the planes and

Let P₁ be the first plane and equation of P₁ is

or x + 3y – 6 = 0

And P₂ 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 P₁ and P₂ is given by

P₁ + λP₂ = 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 (x₁,y₁,z₁) 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 + 9λ² + 9 + λ² – 6λ + 16λ² = 36

⇒ 26λ² + 10 = 36

⇒ 26λ² = 26

⇒ λ² = 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 P₁ be the first plane and equation of P₁ is

or x – y + 2z - 5 = 0

Normal vector to this plane is

And P₂ 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