The image of the point $(-1,3,4)$ in the plane $x-2y=0$ is:

Let the image be $P^{\prime }(x_2,y_2,z_2)$ of point $P(x_1,y_1,z_1)$ in the plane $ax+by+cz+d=0$.Formula: $\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{z_2-z_1}{c}=-2\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}$ Here $(x_1,y_1,z_1)=(-1,3,4)$ and plane is $1x-2y+0z+0=0$.$a=1,b=-2,c=0,d=0$.$\frac{x_2+1}{1}=\frac{y_2-3}{-2}=\frac{z_2-4}{0}=-2\frac{1(-1)-2(3)}{1^2+(-2{)}^2}$$\cdots =-2\frac{-1-6}{5}=-2\frac{-7}{5}=\frac{14}{5}$$x_2+1=14/5⟹x_2=9/5$$y_2-3=-2(14/5)=-28/5⟹y_2=3-28/5=-13/5$$z_2-4=0⟹z_2=4$ Image is $(1.8,-2.6,4)$.Note: The FIITJEE text says "No option matches". However, looking at the provided solution in the prompt, there's a typo in their manual calc but the $z$ -coordinate remains $4$. None of the options fit $(9/5,-13/5,4)$.