The point (2,0,3) lies on the x-z plane as its y-coordinate is 0. Hence it is part of no octant.

The x-coordinate of a point is negative in the octants II, III, VI, VII

For a point in the VIII th octant, only the x-coordinate is positive and the y and z coordinates are negative. Here, (5,-2,-5) lies in the VIII th octant.

The distance of a point P(x,y,z) from the x-axis is
$d=\sqrt{y^2+z^2}=\sqrt{{12}^2+5^2}=\sqrt{144+25}=\sqrt{169}=13units$

The distance of the point P(x,y,z) from the x-y plane is
$d=|z|=9units$

Let the vertices of the triangle be A(0,7,-10), B(1,6,-6) and C(4,9,-6)
$AB=\sqrt{(1-0{)}^2+(6-7{)}^2+(-6-(-10){)}^2}=\sqrt{1^2+1^2+4^2}=\sqrt{18}BC=\sqrt{(4-1{)}^2+(9-6{)}^2+(-6-(-6){)}^2}=\sqrt{3^2+3^2+0^2}=\sqrt{18}AC=\sqrt{(4-0{)}^2+(9-7{)}^2+(-6-(-10){)}^2}=\sqrt{4^2+2^2+4^2}=\sqrt{36}$
$Here,AB=BC\&A{B}^2+B{C}^2=A{C}^2$
Hence, the triangle is isosceles as well as right-angled.

Using the distance formula, the distance between the points (3,0,4) and (-1,3,4) is 
$d=\sqrt{(-1-3{)}^2+(3-0{)}^2+(4-4{)}^2}=\sqrt{4^2+5^2+0^2}=\sqrt{25}=5units$