The distance of the point (1, − 2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x − 2y + z + 12 = 0, is :

The shortest distance between the lines $${x \over 2} = {y \over 2} = {z \over 1}$$ and
$${{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}$$ lies in the interval :

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