The point in which the line x+1−1=y−125=z−72 cuts the surface 11x2−5y2−z2=0 is

Let x+1−1=y−125=z−72 =r any point5 on the line is (−r−1,5r+12,2r+7) for every value of r. if this point lies on the surface 11x2−5y2−z2=0 then 11(−r−1)2−5(5r+12)2+(2r+7)2=0 i.e. 11r2+550r+660=0 i.e. r2+5r+6=0 i.e. (r+3)(r+2)=0 i.e. r=−3,−2 for these two values of r, the two points in which the given line cuts the surface are (2,−3,1) and (1,2,3)