The parametric equation of the line of intersection of the given planes are x-y+z=3 and  2x+y-z=6are

The given planes  are  x-y+z=3and  2x+y-z=6 The line of intersection of two planes is parallel to the vector which is cross product of the normal vectors to the planes hence the vector along the line of intersection of two planes is ijk1−1121−1=i(1-1)−j(−1-2)+k(1+2)This can be simplify as 3j+3kTo get a point on the line of intersection of two planes, substitute  z=1 in the plane equations and then solve for the other two variablesit implies that x-y=2,2x+y=7solving the above two simultaneous equations x=3,y=1Therefore, the parametric form of the equation of the line of intersection of two planes is x=3,y=3t+1,z=3t+1