Given vertices $A(3,-7),\quad B(-1,2),\quad C(4,5)$ Let the orthocenter be $H(\alpha,\beta)$.

Find slope of side $BC$ $m_{BC}=\frac{5-2}{4-(-1)}=\frac{3}{5}$ So the altitude from $A$ is perpendicular to $BC$, hence its slope is $m_{AH}=-\frac{5}{3}$

Equation of altitude through $A(3,-7)$: $y+7=-\frac{5}{3}(x-3)$ $3y+21=-5x+15$ $5x+3y+6=0$

Find slope of side $AC$ $m_{AC}=\frac{5-(-7)}{4-3}=12$ So the altitude from $B$ has slope $m_{BH}=-\frac{1}{12}$

Equation of altitude through $B(-1,2)$: $y-2=-\frac{1}{12}(x+1)$ $12y-24=-x-1$ $x+12y-23=0$

Find intersection of the two altitudes Solve $5x+3y+6=0$ $x+12y-23=0$

From the second equation, $x=23-12y$ Substitute into the first: $5(23-12y)+3y+6=0$ $115-60y+3y+6=0$ $121-57y=0$ $y=\frac{121}{57}$

Then $x=23-12\cdot \frac{121}{57}$ $x=\frac{1311-1452}{57}=-\frac{141}{57}=-\frac{47}{19}$

Hence, $\alpha=-\frac{47}{19},\qquad \beta=\frac{121}{57}$

Compute $9\alpha-6\beta+60$ $9\alpha=9\left(-\frac{47}{19}\right)=-\frac{423}{19}$ $6\beta=6\cdot \frac{121}{57}=\frac{726}{57}=\frac{242}{19}$

Therefore, $9\alpha-6\beta+60=-\frac{423}{19}-\frac{242}{19}+60$ $=-\frac{665}{19}+60$ $=\frac{-665+1140}{19}=\frac{475}{19}=25$

Final answer $9\alpha-6\beta+60=25$ So the correct option is C.