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 $\frac{3}{5}y + \frac{4}{5}z = \frac{6}{5}$where $\frac{3}{5}, \frac{4}{5}$ are the direction cosines of OP.Since direction cosines and direction ratios of the line are proportional, we have $\frac{x_1}{0} = \frac{y_1}{\frac{3}{5}} = \frac{z_1}{\frac{4}{5}} = k$ i.e., $x_1 = 0, y_1 = \frac{3k}{5}, z_1 = \frac{4k}{5}$Substituting these values in the equation of the plane, we get $k = \frac{6}{5}$Hence the foot of perpendicular is $\left(0, \frac{18}{25}, \frac{24}{25}\right)$