Answer :

Given: equation of the planes and

To find: the equation of the plane passing through the line of intersection of the two givenand parallel to the x-axis.

The given equation of planes can be written in the Cartesian form, as shown below

⇒x+y+z-1=0……..(i)

⇒2x+3y-z+4=0……..(ii)

Hence given equation of planes can be written in Cartesian form are

x+y+z-1=0

2x+3y-z+4=0

The equation of a plane passing through the line of intersection of these two planes is,

x+y+z-1+λ(2x+3y-z+4)=0

⇒ x+y+z-1+2 λx+3 λy-z λ+4λ=0

⇒ (1+2λ)x+(1+3λ)y+(1-λ)z+(4λ-1)=0……………(iii)

Now, direction ratios of normal to the plane (iii), are,

a_{1}=1+2λ, b_{1}=1+3λ, c_{1}=1-λ

Now the direction ratios of x-axis are,

a_{2}=1, b_{2}=0, c_{2}=0

Since the required plane (iii) is parallel to x-axis, then

a_{1}a_{2}+b_{1}b_{2}+ c_{1}c_{2}=0

Substituting the corresponding values, we get

⇒ (1+2λ)(1)+(1+3λ)(0)+(1-λ)(0)=0

⇒ 1+2λ+0+0=0

⇒ 2λ=-1

So substituting the value of λ in equation (iii), we get

⇒ -y+3z-6=0

⇒ y-3z+6=0

Is the required equation of the plane, it can also be written as

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