Equation of a plane through the line of intersection of the planes
$x+2y=3, y-2z+1=0 $ is
$\left(x+2y-3\right)+\lambda \left(y-2z+1\right)=0$
$\Rightarrow x+\left(2+\lambda\right)y-2\lambda \left(z\right)-3+\lambda=0 $ (i)
Now, plane(i) is $\bot$ tox + 2y = 3
$\therefore $ Their dot product is zero
i.e. $1+2\left(2+\lambda\right)=0 \Rightarrow \lambda =-\frac{5}{2}$
Thus, required plane is
$x+\left(2-\frac{5}{2}\right) y-2 \times\frac{-5}{2} \left(z\right)-3 -\frac{5}{2}=0$
$\Rightarrow x-\frac{y}{2}+5z -\frac{11}{2} =0$
$\Rightarrow 2x-y+10z-11=0$