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

Given two points,

$\\\vec{A}=\hat{i}-\hat{j}+3\hat{k}\\ \vec{B}=3\left ( \hat{i}+\hat{j}+\hat{k} \right )=3\hat{i}+3\hat{j}+\hat{k}\\ \vec{r}.\left ( 5\hat{i}+2\hat{j}-7\hat{k} \right )+9=0$

Also,

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\\$

Where,

Therefore,

$\left (x\hat{i}+y\hat{j}+z\hat{k} \right ).\left (5\hat{i}+2\hat{j}-7\hat{k} \right )+9=0\\ \\ \Rightarrow 5x + 2y - 7z + 9 = 0$

We must show that the points A and B are equidistant from the plane

5x + 2y - 7z + 9 = 0

We also need to show that the points lie on the opposite side of the plane.

Normal of the plane is, $\vec{N=} 5{i} + 2\hat{j} - 7\hat{k}$

We know, the perpendicular distance of the position vector of a point

$\vec{A} = l\hat{i} + m\hat{j} + n\hat{k} \Rightarrow A(l, m, n)$ to the plane, p: ax + by + cz + d =  0 is given as:

$D=\left | \frac{p(l,m,n)}{\left |\vec{N} \right |} \right |$

Where  $\left |\vec{N} \right |=Normal \: vector\: of\: the\: plane$

$\vec{N} =a\vec{i}+b\vec{j}+c\vec{k}$

Thus, the perpendicular distance of the point $\vec{A} =\vec{i}-\vec{j}+3\vec{k}=A(1,-1,3)$  to the plane 5x + 2y - 7z + 9 = 0 having normal  $\vec{N} =5\vec{i}+2\vec{j}-7\vec{k}$ is given by,

$\left | D_{1} \right |=\left |\frac{5(1)+2(-1)-7(3)+9}{|5\hat{i}+2\hat{j}-7\hat{k}|} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{5-2-21+9}{\sqrt{5^{2}+2^{2}+(-7)^{2}}} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{-9}{\sqrt{25+4+49}} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{9}{\sqrt{78}} \right |\\$

Hence, the perpendicular distance of the point $\vec{B}=3\hat{i}+3\hat{j}+3\hat{k}=B(3,3,3)$ to the plane 5x + 2y - 7z + 9 = 0 having normal  $\vec{N}=5\hat{i}+2\hat{j}-7\hat{k}$

$\left | D_{2} \right |=\left |\frac{5(3)+2(3)-7(3)+9}{|5\hat{i}+2\hat{j}-7\hat{k}|} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{15+6-21+9}{\sqrt{5^{2}+2^{2}+(-7)^{2}}} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{9}{\sqrt{25+4+49}} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{9}{\sqrt{78}} \right |\\$

Therefore, |D1| = |D2|

However, D1 and D2 have different signs.

Therefore, the points A and B will lie on opposite sides of the plane.

Hence, we have successfully shown that the points are equidistant from the plane and lie on opposite sides of the plane.