Since the unknown plane is parallel to the given line (L) : (-1,3,5)+k(2,4,1) then its
normal vector, say n, will be orthogonal to the directing vector u=<2,4,1> of the line (L).
The intersections of the plane 3x-4y+2z=24 with the x-axis and the y--axis are the points
P(8,0,0) and Q(0,-6,0) respectively . The line PQ will lie on the unknown plane. Hence the normal vector n of the unknown plane will also be orthogonal to the vector PQ.
Therefore the vector n is equal to the cross product of vector u and vector PQ.
That is
n=u x (PQ)=<2,4,1> x <-8,-6,0>=<-6,8,20>
Hence the equation of the unknown plane is using the point P (8,0,0)
-6(x-8)+8y+20z=0 or equivalently -3x + 4y + 10z = -24
