The symmetric form of the line of intersection of two planes  x+y+z+1=0,4x+y−2z+2=0 is

The given planes are x+y+z+1=0,4x+y−2z+2=0To get a point on the line, plug in z=0 in the above equations and then solve for x,y  The equations are x+y+1=0,4x+y+2=0A point on the line is -13,-23,0The vector along the line of intersection of two planes is cross product of normal vectors to the planes The vector is n1×n2=ijk11141−2=i(−3)−j(−6)+k(−3)Hence, the direction ratios of the required line are proportional to ⟨1,−2,1⟩Hence, the equation of the required line is x+131=y+23−2=z1