which of the following lines lie on the plane x+ 2y - z + 4= 0?

For line x−11=y−1=z−5−1 point (1, 0, 5) lies on 1-1-ln^n the plane .Also, the vector along the line i^+j^+k^ is perpendicular to the normal i^+2j^-k  to the plane.  For line r→=2i^−j^+4k^+λ(3i^+j^+5k^) point (2, -1,4) lies on the plane and vector 3i^+j^+5k^ is perpendicular to the normal i^+2j^−k^Line x - y + z = 2x + y - z= 0 passes through the origin, which is not on the given plane.