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 $0x+\frac{5}{5}y+0z=-\frac{8}{5}$ where 0, 1, 0 are the direction cosines of OP.Since direction cosines and direction ratios of a line are proportional, then $\frac{x_1}{0}=\frac{y_1}{1}=\frac{z_1}{0}=k$i.e. $x_1 = 0, y_1 = k, z_1 = 0$Substituting these values in the equation of the plane, we get $k=-\frac{8}{5}.$Hence the foot of perpendicular is $\left(0,-\frac{8}{5},0\right).$