Find the equation of the plane passing through the line of intersection of planes 2x - y + z = 3 and 4x- 3y + 5z + 9 = 0 and parallel to the line

` (x+1)/2=(y+3)/4=(z-3)/5`

#### Solution

Given planes are 2x-y+z =3, 4x-3y+5z+9 =0

Equation of required plane passing through their intersection is

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

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

Directio ratios of the normal to the above plane are `2+ 4λ, -1-3λ and 1+ 5λ`

Plane is parallel to the line ` (x+1)/2=(y+3)/4=(z-3)/5`

Direction ratios of line are 2, 4,5

Given that required plane is parallel to given line.

∴normal of the plane is perpendicular to the given line

`(2+4lambda)2+(-1-3lambda)4+(1+5lambda)5=0`

`4+8lambda-4-12lambda+5+25 lambda=0`

`21lambda+5=0`

`therefore lambda=-5/21`

Substituting λ in (1)

∴ Equation of plane is

(2x-y+z-3)-5/21(4x-3y+5z+9)=0

42x-21y+21z-63-20x+15y-25z-45=0

22x-6y-4z-108=0

11x-3y-2z-54=0