Let the coordinates of the foot of the perpendicular P from the origin to the plane be $(x_1, y_1, z_1)$ .Then the direction ratios of the line OP are $x_1, y_1, z_1$ . Writing the equation of the plane in the normal form, we have $\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{1}{\sqrt{3}}$ where $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$ are the direction cosines of OP.Since direction cosines and direction ratios of a line are proportional, then $\frac{x_1}{\frac{1}{\sqrt{3}}}=\frac{z_1}{\frac{1}{\sqrt{3}}}=\frac{y_1}{\frac{1}{\sqrt{3}}}$ i.e., $x_1=\frac{k}{\sqrt{3}},y_1=\frac{k}{\sqrt{3}},z_1=\frac{k}{\sqrt{3}}$Substituting these values in the equation of the plane, we get $k=\frac{1}{\sqrt{3}} .$Hence, the foot of perpendicular is $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3} .\right)$