The plane containing the line  $\frac{x-3}{2}=\frac{y-1}{4}=\frac{z-2}{5}$ is rotated through an angle  $\frac{\pi }{2}$ about this line to contain the point  $(4,3,7).$ The equation of plane before rotation is 

Plane containing the line will be given as $(4x-2y-10)+\lambda (5y-4y+3)=0$
 $\Rightarrow 4x+(5\gamma -2)y-4\lambda z+3\lambda -10=0$ the plane passing through (4,3,7) will give  $\lambda$
 $\Rightarrow 16x+(5\gamma -2)y-4\lambda z+3\lambda -10=0$
 $\Rightarrow -10\lambda =0\Rightarrow \lambda =0$
i.e.,  $4x-2y-10=0$
If this rotated an angle  $\frac{\pi }{2}$ the plane will be  $4x+(5{\lambda }^1-2)y-4{\lambda }^{1}z+(2{\lambda }^1-10)=0$
Also,  $16+(5{\lambda }^1-2)(-2)=0$

$\Rightarrow {\lambda }^1=2 i.e, 4x+8y-8z=4$