The symmetric form of the line x+y-z=1 and 2x-3y+z=2 is

The given planes are  x+y-z=1and 2x-3y+z=2 Plug in z=0 in equations, and then solve for x,yx+y=1,2x-3y=2by solving the above two equations x=1,y=0The direction ratios are proportional to the vector n1×n2=ijk11−12−31=i(−2)−j(3)+k(−5)Therefore, the direction ratios of the required line are proportional to ⟨2,3,5⟩Therefore, the equation of the line in symmetric form is x−12=y3=z5