find the equation of a plane which is at a distance of 3 root 3 units from the origin and the normal to which is equally inclined to the coordinate axes.

Hi,

You know the direction cosine of normal to the plane and u know the point lying on that plane ( say, (3,3,3) in first quadrant) .So, you would get four such planes depending on the quadrant.

thanks................................................................................................

given tht direction ratios are (1,1,1) and perpendicular distance as given 3√3 so,, d/√(1²+1²+1²) =3√3=> d=9 so the required equation is x+y+z=9