
∴ centroid of △ABC=(39+3+5,311+4+13)=(317,328)
Let image of centroid with respect to line mirror is $\begin{aligned}
& (\mathrm{h}, \mathrm{k}) \
& \therefore\left(\frac{\mathrm{k}-\frac{28}{3}}{\mathrm{h}-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1 \
& & 3\left(\frac{\mathrm{h}+\frac{17}{3}}{2}\right)+6 \cdot\left(\frac{\frac{\mathrm{k}+28}{3}}{2}\right)=53
\end{aligned}Solving(1)&(2)weget\mathrm{h}=3, \mathrm{k}=4\therefore \mathrm{h}^2+\mathrm{k}^2+\mathrm{hk}=37$