Step 1:

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

$(2x+y-z-3)+\lambda(5x-3y+4z+9)=0$

$x(2+5\lambda)+y(1-3\lambda)+2(4\lambda-1)+9\lambda-3=0$-----(1)

Step 2:

The plane is parallel to the line

$\large\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}$

$\therefore 2(2+5\lambda)+4(1-3\lambda)+5(4\lambda-1)=0$

(i.e) $18\lambda+3=0$

$\Rightarrow \lambda=-\large\frac{1}{6}$

Step 3:

Put the value of $\lambda$ in emu(1) we obtain

$x(2-\large\frac{5}{6})+y(1+\large\frac{3}{6})+z(-\large\frac{4}{6}-1)-\large\frac{9}{6}-3=0$

$\Rightarrow 7x+9y-10z-27=0$

This is the equation of the required plane.