If an angle between the line, $${{x + 1} \over 2} = {{y - 2} \over 1} = {{z - 3} \over { - 2}}$$ and the plane, $$x - 2y - kz = 3$$ is $${\cos ^{ - 1}}\left( {{{2\sqrt 2 } \over 3}} \right),$$ then a value of k is :

A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(–1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is :

The perpendicular distance from the origin to the plane containing the two lines,

$${{x + 2} \over 3} = {{y - 2} \over 5} = {{z + 5} \over 7}$$ and

$${{x - 1} \over 1} = {{y - 4} \over 4} = {{z + 4} \over 7},$$ is :

If the point (2, $$\alpha $$, $$\beta $$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$$\alpha $$ – 3$$\beta $$ is equal to