Let P(x,y,z) divide the line segment joining A(1,-2,3) and B(3,4,-5) internally in the ratio 2:3. Therefore,

$x=\frac{2\times 3+3\times 1}{2+3}=\frac{9}{5}y=\frac{2\times 4+3\times (-2)}{2+3}=\frac{2}{5}z=\frac{2\times (-5)+3\times 3}{2+3}=\frac{-1}{5}$

$P=(\frac{9}{5},\frac{2}{5},\frac{-1}{5})$

Let P(x,y,z) represent the set of points which equidistant from A(1,2,3) and B(3,2,-1).
Equating AP=BP using the distance formula, we have
$(x-1{)}^2+(y-2{)}^2+(z-3{)}^2=(x-3{)}^2+(y-2{)}^2+(z+1{)}^2\Rightarrow -2x-4y-6z=-6x-4y+2z\Rightarrow 4x-8z=0\Rightarrow x=2z$

Let Q divide PR in the ratio k:1. Equate the x-coordinate of the point of division to 5.

$\therefore \frac{3+9k}{k+1}=53+9k=5k+54k=2k=\frac{1}{2}$
Since it is given that P, Q and R are collinear, we need to calculate k only for any one of the coordinates.

The x and y coordinates are positive and the z coordinate is negative. Hence, it lies in octant V.

For a point in octant VI, the x and z coordinates are negative and the y coordinate is positive.

The y-coordinate of a point is positive in the octants I, II, V and VI

The points which lie on the coordinate axes or coordinate planes are not part of any octant.

The distance of a point P(x,y,z) from the y-axis is
$d=\sqrt{x^2+z^2}=\sqrt{{4.5}^2+6^2}=1.5\times \sqrt{3^2+4^2}=1.5\times 5=7.5units$

Let A=(4,-2,3), B=(5,1,2) and C=(7,7,0).
$AB=\sqrt{(5-4{)}^2+(1-(-2){)}^2+(2-3{)}^2}=\sqrt{1^2+3^2+1^2}=\sqrt{11}$
$BC=\sqrt{(7-5{)}^2+(7-1{)}^2+(0-2{)}^2}=\sqrt{2^2+6^2+2^2}=\sqrt{44}=2\sqrt{11}$
$AC=\sqrt{(7-4{)}^2+(7-(-2){)}^2+(0-3{)}^2}=\sqrt{3^2+9^2+3^2}=\sqrt{99}=3\sqrt{11}$
AC=AB+BC. Hence, A, B and C are collinear. The statement is true.

The distance between the points (-3,7,2) and (2,4,-1), 
$d=\sqrt{(-3-2{)}^2+(7-4{)}^2+(-1-2{)}^2}d=\sqrt{43}$