#### Find the equation of a plane which bisects perpendicularly the line joining A (2, 3, 4) and B (4, 5, 8) at right angles.

Given, there exists a plane which perpendicularly bisects the line joining  A (2, 3, 4) and B (4, 5, 8) at right angles. We must find the equation of this plane.

First, let us find the midpoint of AB.

Since the midpoint of any line is halfway between the two end points,

$Midpoint \: of \: AB=\left ( \frac{2+4}{2},\frac{3+5}{2},\frac{4+8}{2} \right )$

$Midpoint \: of \: AB=\left ( \frac{6}{2},\frac{8}{2},\frac{12}{2} \right )$

= (3, 4, 6).

We can represent this as a position vector,  $\vec{a}=3\hat{i}+4\hat{j}+6\hat{k}$

Next, we must find the normal of the plane, $\vec{n}$

$\vec{n}=\left ( 4-2 \right )\hat{i}+\left ( 5-3 \right )\hat{j}+\left ( 8-4 \right )\hat{k}\\ \Rightarrow \vec{n}=2\hat{i}+2\hat{j}+4\hat{k}$

We know, the equation of the plane which perpendicularly bisects the line joining two given points is

$\left ( \vec{r}-\vec{a} \right )\vec{n}=0$

Where,

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

Substituting the values in the above equation,

$\left (\left (x\hat{i}+y\hat{j}+z\hat{k} \right ) -\left (3\hat{i}+4\hat{j}+6\hat{k} \right ) \right ).\left (2\hat{i}+2\hat{j}+4\hat{k} \right )=0\\ \Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k}-3\hat{i}-4\hat{j}-6\hat{k} \right ).\left (2\hat{i}+2\hat{j}+4\hat{k} \right )=0\\\Rightarrow \left ( x\hat{i}-3\hat{i}+y\hat{j}-4\hat{j}+z\hat{k}-6\hat{k} \right ).\left (2\hat{i}+2\hat{j}+4\hat{k} \right )=0\\\Rightarrow \left ( (x-3)\hat{i}+(y-4)\hat{j}+(z-6)\hat{k} \right ).\left (2\hat{i}+2\hat{j}+4\hat{k} \right )=0\\\Rightarrow 2\left ( x-3 \right )+2(y-4)+4(z-6)=0$

Upon further simplification,

$\Rightarrow 2x-6+2y-8+4z-24=0\\ \Rightarrow 2x+2y+4z-6-8-24=0\\ \Rightarrow 2x+2y+4z-38=0\\ \Rightarrow 2\left ( x+y+2z-19 \right )=0\\ \Rightarrow x+y+2z-19=0\\ \Rightarrow x+y+2z=19$

Therefore, the required equation of the plane is x + y + 2z = 19.