The symmetric form of the line of intersection of planes 3x+2y+z−5=0 and x+y−2z=0

The given planes are x+y−2z=0 and 3x+2y+z−5=0 x=5,y=-5Plug in z=0 to get a point on the line. x+y=0,3x+2y-5=0solve the above two equations, we get x=5,y=-5 A point on the line is 5,-5,0Vector along the line is cross product of two normal vectors to the planes it gives n1×n2=ijk11−2321=i(5)−j(7)+k(−1)Direction ratios of a line of intersection of two planes are proportional to ⟨5,−7,−1⟩Therefore, the equation of the required line is x−55=y+5−7=z−1