Let the coordinates of the foot of perpendicular P from the origin to the given plane be (x₁,y₁,z₁).
We have, 5y + 8 = 0

⇒  0x – 5y + 0z = 8  .... (1)
The direction ratios of the normal are 0, –5 and 0
$\therefore \sqrt{0+{\left(5\right),}^2+0}=5$
Dividing both sides of equation (1) by 5, we obtain
$-\mathrm{y}=\frac{8}{5}$
This equation is of the form ℓx + my + nz = d, where ℓ.m.n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ℓd.md,nd)
Therefore, the given coordinates of the foot of the perpendicular are
$(0-1\left(\frac{8}{5}\right)0)$ i.e. $(0,-\frac{8}{5},0)$