In an isosceles triangle $\mathrm{ABC}$, the vertex $\mathrm{A}$ is (6,1) and the equation of the base B C is $2 x+y=4$. Let the point B lie on the line $x+3 y=7$. If $(\alpha , \beta )$ is the centroid $∆ABC$, then $15 (\alpha +\beta )$ is equal to:

D is foot of perpendicular from A to BC 
$$\begin{gathered} \frac{h-6}{2}=\frac{k-1}{1}=\frac{-(12+1-4)}{2^2+1^2} \\ \frac{h-6}{2}=-\frac{9}{5} \frac{k-1}{1}=-\frac{9}{5} \\ h=\frac{30-18}{5}=\frac{12}{5}; k=-\frac{4}{5} \\ D \left(\frac{12}{5}, -\frac{4}{5}\right) \\ G divides AD in the ratio 2:1 \\ G\left(\alpha ,\beta \right) = \left(\frac{{\displaystyle \frac{24}{5}}+6}{3},\frac{-{\displaystyle \frac{8}{5}}+1}{3}\right) \\ =\left(\frac{54}{15}, -\frac{3}{15}\right) \\ 15 \left(\alpha +\beta \right)=54-3=51 \end{gathered}$$