The required plane Π contains the line L: (-1,1,2)+t(7,6,2) means that Π is perpendicular to the direction vector of the line L, namely vl=<7,6,2>
It is also required that Π be perpendicular to the plane Π 1 : 5y-7z+8=0 means that Π is also perpendicular to the normal vector of the given plane, vp=<0,5,-7>.
Thus the normal vector of the required plane, Π can be obtained by the cross product of vl and vp, or vl x vp: i j k 7 6 2 0 5 -7 =<-42-10, 0+49, 35-0> =<-52, 49, 35> which is the normal vector of Π
Since Π has to contain the line, it must pass through the point (-1,1,2), so the equation of the plane is Π : -52(x-(-1))+49(y-1)+35(z-2)=0 => Π : -52x+49y+35z = 171
Check that normal vector of plane is orthogonal to line direction vector <-52,49,35>.<7,6,2> =-364+294+70 =0 ok
