The symmetric form of the line of intersection of two planes 3x−y=6,y+2z−1=0

The given planes are 3x−y=6,y+2z−1=0 Plug in z=0 to get a point on the line. 3x-y=6,y=1solve the above two equations, we get x=73,y=1 A point on the line is 73,1,0Vector along the line is cross product of two normal vectors to the planes it gives n1×n2=ijk3-10012=i(-2)−j(6)+k(3)Direction ratios of a line of intersection of two planes are proportional to ⟨2,6,−3⟩Therefore, the equation of the required line is x−732=y-16=z-3