Show that the points $(\hat{i}-\hat{j}+3 \hat{k})$ and $3(\hat{i}+\hat{j}+\hat{k})$ are equidistant from the plane $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$ and lie on opposite sides of it.

(N/A) Let the given points be $\vec{a} = \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{b} = 3\hat{i}+3\hat{j}+3\hat{k}$. The equation of the plane is $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$.
The distance $d$ of a point $\vec{p}$ from the plane $\vec{r} \cdot \vec{n} + d_0 = 0$ is given by $d = \frac{|\vec{p} \cdot \vec{n} + d_0|}{|\vec{n}|}$.
For point $\vec{a}$:
$d_1 = \frac{|(1)(5) + (-1)(2) + (3)(-7) + 9|}{\sqrt{5^2 + 2^2 + (-7)^2}} = \frac{|5 - 2 - 21 + 9|}{\sqrt{25 + 4 + 49}} = \frac{|-9|}{\sqrt{78}} = \frac{9}{\sqrt{78}}$.
For point $\vec{b}$:
$d_2 = \frac{|(3)(5) + (3)(2) + (3)(-7) + 9|}{\sqrt{5^2 + 2^2 + (-7)^2}} = \frac{|15 + 6 - 21 + 9|}{\sqrt{78}} = \frac{|9|}{\sqrt{78}} = \frac{9}{\sqrt{78}}$.
Since $d_1 = d_2$,the points are equidistant.
To check the sides,we evaluate $f(\vec{r}) = \vec{r} \cdot \vec{n} + d_0$ for both points:
$f(\vec{a}) = 5 - 2 - 21 + 9 = -9 < 0$.
$f(\vec{b}) = 15 + 6 - 21 + 9 = 9 > 0$.
Since the values have opposite signs,the points lie on opposite sides of the plane.