Let line $\mathrm{AB}$ be $\mathrm{x}+3 \mathrm{y}=7$ and point $\mathrm{P}$ be $(3,8)$. Let $\mathrm{Q}$ (h, k) be the image of point $P(3,8)$ in the line $x+3 y=7$. Since line $\mathrm{AB}$ is a mirror.
(1) Point $\mathrm{P} \& \mathrm{Q}$ are at equal distance from line $\mathrm{AB}$ i.e. $P R=Q R$ i.e. $R$ is the mid-point of $P Q$.
(2) Image is formed perpendicular to mirror i.e. line $P Q$ is perpendicular to line $\mathrm{AB}$.
Since $\mathrm{R}$ is the midpoint of $\mathrm{PQ}$.
Mid point of $\mathrm{PQ}$ joining $(3,8)$ and $(\mathrm{h}, \mathrm{k})$ is $\left(\frac{\mathrm{h}+3}{2}, \frac{\mathrm{h}+8}{2}\right)$
Coordinate of point $\mathrm{R}=\left(\frac{\mathrm{h}+3}{2}, \frac{\mathrm{h}+8}{2}\right)$
Since point $\mathrm{R}$ lies on the line $\mathrm{AB}$.
Therefore, $\left(\frac{3+\mathrm{h}}{2}\right)+3\left(\frac{8+\mathrm{k}}{2}\right)=7$
$$
\Rightarrow \mathrm{h}+3 \mathrm{k}=-13.....(i)
$$
Also, $\mathrm{PQ}$ is perpendicular to $\mathrm{AB}$
Therefore, slope of $\mathrm{PQ} \times$ Slope of $\mathrm{AB}=-1$
$\Rightarrow$ Slope of $\mathrm{PQ} \times\left(-\frac{1}{3}\right)=-1$
$\Rightarrow$ Slope of $\mathrm{PQ}=3$
Now, $\mathrm{PQ}$ is the line joining $\mathrm{P}(3,8)$ \& $\mathrm{Q}(\mathrm{h}, \mathrm{k})$.
$\therefore$ Slope of $\mathrm{PQ}=3=\frac{\mathrm{k}-8}{\mathrm{~h}-3} \Rightarrow 3 \mathrm{~h}-\mathrm{k}=1$.....(ii)
Solving eqn. (i) \& (ii), we get
$$
\begin{aligned}
& \mathrm{h}=-1 \& \mathrm{k}=-4 \\
& \therefore \text { Image is } \mathrm{Q}(-1,-4)=(\alpha, \beta) \Rightarrow \alpha=-1, \beta=-4 \\
& \alpha+\beta=-1+(-4)=-5
\end{aligned}
$$