Step 1:
The given equation of the plane is $x+y+z=1$------(1)
Hence the direction cosines of the normal are $(1,1,1)$
Therefore $\sqrt{1^2+1^2+1^2}=\sqrt 3$
Divide both sides of the equation by $\sqrt 3$
$\large\frac{1}{\sqrt 3}$$x+\large\frac{1}{\sqrt 3}$$y+\large\frac{1}{\sqrt 3}$$z=\large\frac{1}{\sqrt 3}$
Step 2:
Therefore the coordinates of the foot of the perpendicular are $(ld,md,nd)$
(i.e)$\big(\large\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}.\frac{1}{\sqrt 3}\big)$
$\Rightarrow (\large\frac{1}{ 3},\large\frac{1}{ 3},\large\frac{1}{ 3})$
Hence the coordinates are $(\large\frac{1}{ 3},\large\frac{1}{ 3},\large\frac{1}{ 3})$