# Find the equation

Given: equation of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 =0

To find: the equation of the plane passing through the line of intersection of the two given and parallel to the line

Equations of the planes are

2x + y – z -3=0 and 5x – 3y + 4z + 9 =0

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

2x + y – z - 3 +λ(5x – 3y + 4z + 9)=0

2x + y – z - 3 +5λx-3λy+4z λ+9λ=0

(2+5λ)x+(1-3λ)y+(4λ-1)z+(9λ-3)=0……………(iii)

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

a1=2+5λ, b1=1-3λ, c1=4λ-1

Now plane (iii) is parallel to line

Now the direction ratios of this line are,

a2=2, b2=4, c2=5

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

a1a2+b1b2+ c1c2=0

Substituting the corresponding values, we get

(2+5λ)(2)+(1-3λ)(4)+( 4λ-1)(5)=0

4+10λ+4-12 λ+20 λ-5=0

3+18λ=0

18λ=-3

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

Multiplying both sides with 6 we get

7x+9y-10z-27=0

7x+9y-10z=27

Is the required equation of the plane

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