Sort the points by linear distance in a 3D space

05AB1E, 4 bytes

ΣαnO

Try it online!

Explanation

Σ        # sort by
   O     # sum of
  n      # square of
 α       # absolute difference between current value and second input

JavaScript (ES6), 71 bytes

(b,a,g=a=>a.reduce((d,c,i)=>d+(c-=b[i])*c,0))=>a.sort((b,a)=>g(b)-g(a))

Haskell, 54 52 bytes

import Data.List
f o=sortOn(sum.map(^2).zipWith(-)o)

Try it online!

I don't need the size of the space. sum.map(^2).zipWith(-)o computes the distance from a point to o : (xo-xp)^2+(yo-yp)^2+(zo-zp)^2. The points are simply sorted on the distance to o.

EDIT: "if you don't need it, don't take it" saved 2 bytes.