The equation of a plane, containing the line of intersection of the planes $$2 x-y-4=0$$ and $$y+2 z-4=0$$ and passing through the point $$(2,1,0)$$, is

The foot of the perpendicular drawn from the origin to the plane is $$(4,-2,5)$$, then the Cartesian equation of the plane is

A vector $$\overrightarrow{\mathrm{n}}$$ is inclined to $$\mathrm{X}$$-axis at $$45^{\circ}$$, $$\mathrm{Y}$$-axis at $$60^{\circ}$$ and at an acute angle to Z-axis If $$\overrightarrow{\mathrm{n}}$$ is normal to a plane passing through the point $$(-\sqrt{2}, 1,1)$$, then equation of the plane is

If the Cartesian equation of a line is $$6 x-2=3 y+1=2 z-2$$, then the vector equation of the line is