If the line, $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z + 4} \over 3}\,$$ lies in the planes, $$lx+my-z=9,$$ then $${l^2} + {m^2}$$ is equal to :

The distance of the point $$(1, 0, 2)$$ from the point of intersection of the line $${{x - 2} \over 3} = {{y + 1} \over 4} = {{z - 2} \over {12}}$$ and the plane $$x - y + z = 16,$$ is :

The equation of the plane containing the line $$2x-5y+z=3; x+y+4z=5,$$ and parallel to the plane, $$x+3y+6z=1,$$ is :

The image of the line $${{x - 1} \over 3} = {{y - 3} \over 1} = {{z - 4} \over { - 5}}\,$$ in the plane $$2x-y+z+3=0$$ is the line :