The equation of the plane containing the line of intersection of the planes $2x-y=0\text{ and }y-3z=0$  and perpendicular to the plane $4x+5y-3z-8=0$ 

Equation of plane containing the line of intersection of planes is,  $(2x-y)+\lambda (y-3z)=0$ ... (i)

Also, plane (i) is perpendicular to  $4x+5y-3z-8=0$

$\begin{array}{l}\therefore 4\left(2\right)+5(\lambda -1)-3(-3\lambda )=0\\ \Rightarrow 14\lambda =-3\Rightarrow \lambda =-\frac{3}{14}\end{array}$

Put the value of $\lambda$ in (i), we get $28x-17y+9z=0,$ , which is the required plane.