Prove that the lines (x-a)/a'=(y-b)/b'=(z-c)/c' and (x-a')/a=(y-b')/b=(z-c')/c intersect and find the coordinates of the point of intersection and the equation of the plane in which they lie.
If (x-a)/a'=(y-b)/b'=(z-c)/c'=s, then (x,y,z) = (a,b,c) + s(a',b',c'). Similarly, the equation of the other line can be written in vector form as (x,y,z) = (a',b',c') + t(a,b,c). It should be obvious from that that you can get a point on both lines by putting s=t=1.
Also, the directions given by the vectors (a,b,c) and (a',b',c') both lie in the plane containing the two lines. So you can get a vector normal to the plane by taking the cross product of those two vectors. From that, you can find the equation of the plane.