Equation of the planes are 2x+y−z=3 and 5x−3y+4z+9=0.
The equation of the plane passing through the line of intersection of these planes is
(2x+y−z−3)+λ(5x−3y+4z+9)=0
x(2+5λ)+y(1−3λ)+z(4λ−1)+9λ−3=0 .......(1)
The plane is parallel to the line :
x−12=y−34=z−55
Therefore,
2(2+5λ)+4(1−3λ)+5(4λ−1)=0
18λ+3=0
λ=−16
Putting this value in equation 1, we get,
x(2−56)+y(1+36)+z(−46−1)−96−3=0
7x+9y−10z−27=0
This is the equation of the required plane.