The equation of line parallel to the plane x+y+z=2 and intersecting the line x+y−z=0=x+2y−3z+5 at the point  -4,1,3 is

The equation of plane is x+y−z+λ(x+2y−3z+5)=0It is passing through the point -4,1,3It implies that−4+1−3+λ(−4+2−9+5)=0⇒λ=−1Hence, the equation of plane is  y−2z+5=0The required line is parallel to both planes y−2z+5=0 and x+y+z=2Normal vectors to above planes aren1=i+j+k,n2=j−2k  The vector along the line is perpendicular to both the vectors, so it is along  n1×n2Hence,n1×n2=ijk11101−2=−3i+2j+kHence, the direction ratios of the required line are ⟨−3,2,1⟩ Therefore, the equation of the required line is  x+4−3=y−12=z−31