Find the equation of a plane which is at a distance $3\sqrt 3$ units from origin and the normal to which is equally inclined to coordinate axis.

Since, normal to the plane is equally inclined to the coordinate axis.

Therefore, $\cos \alpha = \cos \beta = \cos \gamma = \frac{1}{{\sqrt 3 }}$

So, the normal is $\overrightarrow {\rm{N}} = \frac{1}{{\sqrt 3 }}\widehat {\rm{i}} + \frac{1}{{\sqrt 3 }}\widehat {\rm{j}} + \frac{1}{{\sqrt 3 }}\widehat {\rm{k}}$

and plane is at a distance of $3\sqrt 3$ units from origin.

The equation of plane is $\overrightarrow {\rm{r}} \cdot \widehat {\rm{N}} = 3\sqrt 3$
[since, vector equation of the plane at a distance $p$ from the origin is

$\overrightarrow {\rm{r}} \cdot \widehat {\rm{N}} = p]$

$\Rightarrow$ $(x\widehat {\rm{i}} + y\widehat {\rm{j}} + z\widehat {\rm{k}}) \cdot \frac{{\left( {\frac{1}{{\sqrt 3 }}\widehat {\rm{i}} + \frac{1}{{\sqrt 3 }}\widehat {\rm{j}} + \frac{1}{{\sqrt 3 }}\widehat {\rm{k}}} \right)}}{1} = 3\sqrt 3$

$\Rightarrow$ $\frac{x}{{\sqrt 3 }} + \frac{y}{{\sqrt 3 }} + \frac{z}{{\sqrt 3 }} = 3\sqrt 3$

$\therefore$ $x + y + z = 3\sqrt 3 \cdot \sqrt 3 = 9$

So, the required equation of plane is $x + y + z = 9$.