Given equations of the given plane are $x-2y+3z=0$..........(i) and
$2x+3y-4z-5=0$.........(ii)
Equation of the plane through the line of intersection of (i) and (ii) is given by
$x-2y+3z+\lambda(2x+3y-4z-5)=0$.
$=x(1+2\lambda)+(y(-2+3\lambda)+z(3-4\lambda)-5\lambda=0$...........(iii)
The equation of the $x-2y+3z=0$ in its new position given by (iii).
But the new position is got by rotating (i) by $90^{\circ}$
$\therefore$ The angle between (i) and (iii) is $90^{\circ}$
$\Rightarrow\:(1,-2,3).(1+2\lambda,\:\:-2+3\lambda,\:\:3-4\lambda)=0$
$\Rightarrow\:1+2\lambda+4-6\lambda+9-12\lambda=0$
$\Rightarrow\:\lambda=\large\frac{7}{8}$
$\therefore$ By substituting $\lambda$ in the required eqn. of the plane (iii) it becomes
$22x+5y-4z+35=0$