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 $\frac{2}{\sqrt{29}}x+\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{12}{\sqrt{29}}$where $\frac{2}{\sqrt{29}},\frac{3}{\sqrt{29}},\frac{4}{\sqrt{29}}$ are the direction cosines of OP.Since direction cosines and direction ratios of a line are proportional, we have $\frac{x_1}{\frac{2}{\sqrt{29}}}=\frac{y_1}{\frac{3}{\sqrt{29}}}=\frac{z_1}{\frac{4}{\sqrt{29}}}=k$ i.e., $x_1=\frac{2k}{\sqrt{29}},y_1=\frac{3k}{\sqrt{29}}$, $z_1=\frac{4k}{\sqrt{29}}$Substituting these values in the equation of plane, we get $k=\frac{12}{\sqrt{29}}$Hence foot of perpendicular is $\left(\frac{24}{29},\frac{36}{29},\frac{48}{29}\right)$