If from a point P(a, b, c), perpendicular PA and PB are drawn to YZ and ZX-planes respectively, then the equation of the plane OAB is

A line with positive direction cosines passes through the point P(2, $$-$$1, 2) and makes equal angle with co-ordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals.

The plane lx + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle $$\alpha$$. The equation changes to

The equation of the plane through the point $$(2, - 1, - 3)$$ and parallel to the lines

$${{x - 1} \over 2} = {{y + 2} \over 3} = {z \over { - 4}}$$ and $${x \over 2} = {{y - 1} \over { - 3}} = {{z - 2} \over 2}$$ is