Find the line of intersection between the two planes z-x-y=0 and z-2x+y=0 .

First, know that the vector that is in the direction of the line of intersection (r) is the cross product between the normal vectors of the planes since it is perpendicular to each of them. A plane in the form ax+by+cz=0 has a normal vector n=lta,b,cgt .
Let z-x-y=0 be plane 1 and z-2x+y=0 be plane 2 . Then the two normal vectors are n_1=lt-1,-1,1gt and n_2=lt-2,1,1gt .
r=n_1 xx n_2=(-1*-1)i-(-1+2)j+(-1-2)k=-2i-j-3k=lt-2,-1,-3gt
A line has the parametric equation
L(t)=(a,b,c)-t*r
Where point (a,b,c) is any point on the line in the direction of r. So all we need now is a point on the line. The vector r has a z component of -3 which means at some value of t it must go through the plane z=0 . Therefore, we can set z=0 in the plane 1 and 2 equations then solve for the x and y coordinates.
(1):-> 0-x-y=0
(2):-> 0-2x+y=0
Solving these equations gives x=0 and y=0 . So a point that the plane goes through is (0,0,0) . Then an equation has for the line must be
L(t)=(0,0,0)-t*r
L(t)=-t*lt-2,-1,-3gt
L(t)=t*lt2,1,3gt
The explicit parametric equations are:
X(t)=2t
Y(t)=t
Z(t)=3t