Let C(x,y,z) be the centroid of the triangle whose vertices are (0,3,5), (1,4,-2) and (2,-1,3)

$x=\frac{0+1+2}{3}=\frac{3}{3}=1y=\frac{3+4-1}{3}=\frac{6}{3}=2z=\frac{5-2+3}{2}=\frac{6}{3}=2$

The centroid of the triangle is (1,2,2)

Let D(x,y,z) be the fourth vertex of the parallelogram.
The diagonals of a parallelogram bisect each other.
Therefore, the midpoints of AC and BD are the same. Let this be O.

$O=(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2})O=(1,0,2)$

Now, O is the midpoint of BD
$\therefore \frac{1+x}{2}=1\Rightarrow x=1\frac{2+y}{2}=0\Rightarrow y=-2\frac{4+z}{2}=2\Rightarrow z=0$

D=(1,0,2)

In the VII th octant, all three coordinates of a point are negative. Here, only the point (-2,-3,-10) lies in the seventh quadrant.

For a point on the x-y plane, the z-coordinate is zero and the x and y coordinates are non-zero. Here, (1,5,0) lies on the x-y plane.

For a point on the x-z plane, the y-coordinate is zero and the x and z coordinates are non-zero. Here, (3,0,-6) lies on the x-z plane.

The distance of a point P(x,y,z) from the z-axis is
$d=\sqrt{x^2+y^2}=\sqrt{{15}^2+8^2}=\sqrt{225+64}=\sqrt{289}=17units$

Let P(x,y,z) be the required point.
According to the given condition
$\sqrt{(x-0{)}^2+(y-0{)}^2+(z-0{)}^2}=3\Rightarrow x^2+y^2+z^2=9$

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

$x=\frac{2\times 3+(-3)\times 1}{2+(-3)}=-3y=\frac{2\times 4+(-3)\times (-2)}{2+(-3)}=-14z=\frac{2\times (-5)+(-3)\times 3}{2+(-3)}=19$

$P=(-3,-14,19)$

Let the YZ-plane divide the line segment joining the points (4,8,10) and (6,10,-8) in the ratio k:1 at the point P.
The x-coordinate of the point P is given by
$x=\frac{4+6k}{k+1}$
Since P lies on the YZ plane, x=0
$\Rightarrow 4+6k=0\Rightarrow k=\frac{-2}{3}$

Hence, the points are divided by the YZ-plane in the ratio 2:3 externally