The equation of the line of intersection of planes 4x+4y−5z=12 and 8x+12y−13z=32 can be written as

Given equation of planes are  4x+4y−5z=12−−−−−(i) And 8x+12y−13z=32−−−−−(ii) Let DR's of required line be (l,m,n)From Eq’s (i) and (ii) we get 4l+4m−5n=0 And 8l+12m−13n=0⇒l8=m12=n16⇒l2=m3=n4 Now we take intersection point with z=0 is given by 4x+4y=12−−−−(iii) And 8x+12y=32−−−−(iv) On solving Equations (i) and (ii) we get the point (1,2,0) Required line is x−12=y−23=z4