In the following case,determine the direction cosines of the normal to the plane and the distance from the origin: $x+y+z=1$

The given equation of the plane is $x+y+z=1$ ............$(1)$
The direction ratios of the normal to the plane are $a=1, b=1, c=1$.
The magnitude of the normal vector is $\sqrt{a^{2}+b^{2}+c^{2}} = \sqrt{(1)^{2}+(1)^{2}+(1)^{2}} = \sqrt{3}$.
To find the direction cosines,we divide the equation $(1)$ by $\sqrt{3}$:
$\frac{1}{\sqrt{3}} x + \frac{1}{\sqrt{3}} y + \frac{1}{\sqrt{3}} z = \frac{1}{\sqrt{3}}$ ..........$(2)$
This equation is in the normal form $lx + my + nz = d$,where $l, m, n$ are the direction cosines of the normal to the plane and $d$ is the distance of the plane from the origin.
Comparing equation $(2)$ with the normal form,the direction cosines of the normal are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ and the distance from the origin is $\frac{1}{\sqrt{3}}$ units.