ABC is a triangle in a plane with vertices

A(2, 3, 5), B(−1, 3, 2) and C($$\lambda $$, 5, $$\mu $$).

If the median through A is equally inclined to the coordinate axes, then the value of ($$\lambda $$³ + $$\mu $$³ + 5) is :

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 :