3-D geometry can have 3 coordinates each with positive or negative value.
Total octants = 2*2*2 = 8.

The coordinates of a point dividing line segment joining (x₁, y₁, z₁) and (x₂, y₂, z₂) internally in ratio m: n is \((\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})\).
Let ratio be k : 1.So, coordinates of a point dividing line segment joining (1, 2, 3) and (4, 5, 6) internally in ratio k: 1 is \((\frac{k*4+1*1}{k+1},\frac{k*5+1*2}{k+1},\frac{k*6+1*3}{k+1})\)
=> \((\frac{k*4+1*1}{k+1},\frac{k*5+1*2}{k+1},\frac{k*6+1*3}{k+1})\) is same as (3, 4, 5).
=> (4k+1)/(k+1) = 3
=> 4k+1 = 3k+3
=> k = 2
So, ratio is 2:1.

We know, distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
Distance AB = \(\sqrt{(1-3)^2+(2-1)^2+(3-2)^2} = \sqrt{6}\)
Distance BC = \(\sqrt{(3-2)^2+(1-3)^2+(2-1)^2} = \sqrt{6}\)
Distance CA = \(\sqrt{(2-1)^2+(3-2)^2+(1-3)^2} = \sqrt{6}\)
Since AB=BC=CA so, it forms equilateral triangle.

In 3d coordinate system, a point with y-coordinate zero and x & z coordinate having non-zero value must lie on X-Z plane. So, (3, 0, 4) lie on X-Z plane.

Since point with all negative coordinates i.e. of form (-, -, -) lie in 7^th octant.
So, (-1, – 5, -7) lies in 7^th octant.

We know, distance of a point from X-Y plane is equal to value of its z-coordinate.
So, distance of point (2, 3, 5) from X-Y plane is 5 units.

Points which trisect line divides it into 2:1 and 1:2.
The coordinates of a point dividing line segment joining (x₁, y₁, z₁) and (x₂, y₂, z₂) internally in ratio m : n is \((\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})\).
For 1:2, coordinates of point are \((\frac{1*13+2*4}{1+2},\frac{1*27+2*9}{1+2},\frac{-4+2*8}{1+2})\) = (7, 15, 4)
For 2:1, coordinates of point are \((\frac{2*13+1*4}{1+2},\frac{2*27+1*9}{1+2},\frac{-8+1*8}{1+2})\) = (10, 21, 0)

We know, distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\).
So, distance between two points (5, 6, 7) and (2, 6, 3) will be \(\sqrt{(5-2)^2+(6-6)^2+(7-3)^2}\) = \(\sqrt{(3)^2+(4)^2}\) = 5 units.

We know, distance of a point from X-Z plane is equal to value of its y-coordinate.
So, distance of point (2, 3, 5) from X-Z plane is 3 units.

If coordinates of vertices of a triangle are (x₁, y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃) coordinates of centroid of triangle are ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3)
So, coordinates of centroid of given triangle are ((7+5+9)/3, (6+4+5)/3, (4+6+8)/3) = (7, 5, 3).