Let $Q$ be image of the point $P(1,3,4)$ in the given plane, then $P Q$ is normal to the plane. The d.r.'s of $P Q$ are $2,-1,1$
Since $P Q$ passes through $(1,3,4)$ and has d.r's $2,1,-1$;
therefore, equation of plane is $\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=r$
$\therefore \quad x=2 r+1, y=-r+3, z=r+4$
$\therefore \quad x=2 r+1, y=-r+3, z=r+4$

So, co-ordinates of $Q$ be
$(2 r+1,-r+3, r+4)$
Let $R$ be the mid point of $P Q$, then co-ordiantes of $R$ are $\left(\frac{2 r+1+1}{2}, \frac{-r+3+3}{2}, \frac{r+4+4}{2}\right)$
i.e., $\left(r+1, \frac{-r+6}{2}, \frac{r+8}{2}\right)$

Since $R$ lies on the plane
$\therefore 2(r+1)-\left(\frac{-r+6}{2}\right)+\left(\frac{r+8}{2}\right)+3=0 \Rightarrow r=-2$

So, co-ordinates of $Q$ are $(-3,5,2)$.
Trick : From option (a), mid point of $(-3,5,2)$ and $(1,3,4)$ satisfies the equation of plane $2 x-y+z+3=0$.