If the distance between the plane $$\alpha x - 2y + z = k$$ and the plane containing the lines $${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}$$ and $${{x - 2} \over 3} = {{y - 3} \over 4} = {{z - 4} \over 5}$$ is $$\sqrt 6 $$, then $$|k|$$ is

The angle between a normal to the plane $$2x - y + 2z - 1 = 0$$ and the X-axis is

The equation of the plane through the intersection of the planes x + y + z = 1 and 2x + 3y $$-$$ z + 4 = 0 and parallel to the x-axis is

The line $$x - 2y + 4z + 4 = 0$$, $$x + y + z - 8 = 0$$ intersect the plane $$x - y + 2z + 1 = 0$$ at the point