Suppose the line L is contained in the given plane. Since (x1, y1, z1) lies on L and L is contained in E = 0 we have
ax1 + by1 + cz1 + d = 0
Also the normal (a,b,c) of E = 0 is perpendicular to the line L. This implies
(a,b,c).(l,m,n) = 0
al + bm + cn = 0
Conversely, assume that
ax1 + by1 + cz1 + d = 0 ...(1)
and al + bm + cn = 0 ...(2)
Since one of a, b and c is not zero, we have that the vector (a, b, c) is normal to the plane E ≡ ax + by + cz + d = 0. Also from Eqs. (1) and (2), we have that (x1, y1, z1) lies in E = 0 and al + bm + cn = 0 which implies that (l,m,n) is perpendicular to the normal (a,b,c) or E = 0. Hence, L must lie in the plane E = 0.