An example of using planes to find the equations of a line in three - space.

The first plane has equation 2 x - y + 3 z = 0, so A1 = 2, B1 = -1, C1 = 3, and D1 = 0;
The second plane has equation x + 4 y - 2 z + 1 = 0, so A2 = 1, and so on.

Here we graph them by solving for z as a function of x and y in each case (what modification would you have to make if the z coefficient were 0 in one or both equations?):

(* At the moment, this graphing only works when C1 and C2 are not 0. *)
		A1 = 2; B1 = -1; C1 = 3; D1 = 0;
		plane1 = Plot3D[(-A1/C1)x+(-B1/C1)y -D1/C1,{x,-10,10},{y,-10,10},Mesh->False,FillingStyle->Opacity[.25]];
		A2 = 1; B2 = 4; C2 = -2; D2 = 1;
		plane2 = Plot3D[(-A2/C2)x+(-B2/C2)y -D2/C2,{x,-10,10},{y,-10,10},Mesh->False,FillingStyle->Opacity[.25]];

	scale = {-4,4};
	Show[plane1,plane2,PlotRange->{scale,scale,scale},Axes->True,AxesOrigin->{0,0,0},Boxed->False,AspectRatio->1]

Find equations for line of intersection?  We use the standard algebra on systems of equations to "solve" for each of other variables in terms of one chosen variable.  Then we set that variable as a parameter (we typically call this parameter time) t and write all three variables in terms of time.  Then we graph along with the two planes.  
      
      Step 1 :  (Eq 1) + (-2) (Eq 2) yields - 9 y + 7 z = 2, or y = (7/9) z - (2/9).
      Step 2 :  4 (Eq 1) + (Eq 2) yields 9 x + 10 z = -1, or x = -(10/9) z - (1/9).
      
Since both x and y are written in terms of z, let's let z = t.  
Then the equations for the line of intersection of these two planes in three - space are 
				x = -(10/9) t - (1/9),   y = (7/9) t - (2/9),   and   z = t;
Is this the right line?  Let' s graph it :

line = ParametricPlot3D[{-(10/9)t-(1/9), (7/9)t-(2/9), t},{t,-10,10}, PlotStyle-> {Thick,Red}];

Let' s show the planes and the line of intersection on the same graph :

Show[plane1,plane2,line,PlotRange->{scale,scale,scale},Axes->True,AxesOrigin->{0,0,0},Boxed->False,AspectRatio->1]