The distance of the point $$(1,-5,9)$$ from the plane $$x-y+z=5$$ measured along the line $$x=y=z$$ 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 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 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 :