If α ,   β is the orthocenter of the triangle A B C with vertices A 3 ,   – 7 ,…

The required diagram will be: 

Equation of $AD:y+7=\frac{-5}{3}\left(x-3\right)$

$\Rightarrow 3y+21=-5x+15$

$\Rightarrow 5x+3y+6=0 ......\left(\mathrm{i}\right)$

Equation of $BE:y-2=\frac{-1}{12}\left(x+1\right)$

$\Rightarrow 12y-24=–x-1$

$\Rightarrow x=23-12y....\left(ii\right)$

by $\left(\mathrm{ii}\right)$ $115-60y+3y+6=0$

$\Rightarrow 57y=121$

$y=\frac{121}{57}, x=23-12\times \frac{121}{57}$

$\therefore 9\alpha -6\beta +60=9\times 23-108\times \frac{121}{57}-6\times \frac{121}{57}+60$

$=207-242+60=25$

Hence this is the required option.