Show that the points $(\widehat {\rm{i}} - \widehat {\rm{j}} + 3\widehat {\rm{k}})$ and $3(\widehat {\rm{i}} + \widehat {\rm{j}} + \widehat {\rm{k}})$ are equidistant from the plane $\overrightarrow {\rm{r}} \cdot (5\widehat {\rm{i}} + 2\widehat {\rm{j}} - 7\widehat {\rm{k}}) + 9 = 0$ and lies on opposite side of it.

To show that these given points

$(\widehat {\rm{i}} - \widehat {\rm{j}} + 3\widehat {\rm{k}})$ and $3(\widehat {\rm{i}} + \widehat {\rm{j}} + \widehat {\rm{k}})$

are equidistant from the plane $\overrightarrow {\rm{r}} \cdot (5\widehat {\rm{i}} + 2\widehat {\rm{j}} - 7\widehat {\rm{k}}) + 9 = 0$,

we first find out the mid- point of the points which is $2\widehat {\rm{i}} + \widehat {\rm{j}} + 3\widehat {\rm{k}}$.

On substituting $\overrightarrow {\rm{r}}$ by the mid-point in plane,

we get ${\rm{LHS}} = (2\widehat {\rm{i}} + \widehat {\rm{j}} + 3\widehat {\rm{k}}) \cdot (5\widehat {\rm{i}} + 2\widehat {\rm{j}} - 7\widehat {\rm{k}}) + 9$

$= 10 + 2 - 21 + 9 = 0$

$= {\rm{RHS}}$

Hence, the two points lie on opposite sides of the plane are equidistant from the plane.