The points $C$ between $A$ and $B$ can be parametrized by $ C=At+B(1-t)\tag{1} $ where $0< t< 1$
Therefore, we wish to solve for $t\in(0,1)$ so that $ \begin{align} |C-A|&=2|C-B|\\ |At+B(1-t)-A|&=2|At+B(1-t)-B|\\ |(B-A)(1-t)|&=2|(A-B)t|\\ 1-t&=2t\\ &t=\tfrac13\tag{2} \end{align} $ Thus, plugging $(2)$ into $(1)$ yields $ \begin{align} C &=\tfrac13A+\tfrac23B\\ &=\tfrac13(2, -3, 1)+\tfrac23(8, 9, -5)\\ &=(6,5,-3)\tag{3} \end{align} $
Determining if $C$ is between $A$ and $B$:
In general, $C$ is between $A$ and $B$ when $ |B-A|=|B-C|+|C-A|\tag{4} $ Equation $(4)$ is the extreme case of the triangle inequality, which says $ |B-A|\le|B-C|+|C-A|\tag{5} $
Check:
Let's check if $C=(6, 7, 3)$ is between $A=(2, -3, 1)$ and $B=(8, 9, -5)$: $ |A-B|=|(-6,-12,6)|=6\sqrt{6} $ but $ |A-C|+|C-B|=|(-4,-10,-2)|+|(-2,-2,8)|=2\sqrt{30}+6\sqrt{2} $ Numerically, $6\sqrt{6}\approx14.70$ and $2\sqrt{30}+6\sqrt{2}\approx19.44$. Thus, $ (6, 7, 3)\text{ is not between }(2, -3, 1)\text{ and }(8, 9, -5)\tag{6} $ Let's check if $C=(6, 5, -3)$ is between $A=(2, -3, 1)$ and $B=(8, 9, -5)$: $ |A-B|=|(-6,-12,6)|=6\sqrt{6} $ and $ |A-C|+|C-B|=|(-4,-8,4)|+|(-2,-4,2)|=4\sqrt{6}+2\sqrt{6}=6\sqrt{6} $ Thus, $ (6, 5, -3)\text{ is between }(2, -3, 1)\text{ and }(8, 9, -5)\tag{7} $
Extension:
Given this question, one might wonder what is the set of all points $C$ so that $ |C-A|=2|C-B|\tag{8} $ Squaring $(8)$ yields $ C\cdot C-2A\cdot C+A\cdot A=4C\cdot C-8B\cdot C+4B\cdot B $ $ 0=3C\cdot C-2(4B-A)\cdot C+4B\cdot B-A\cdot A $ Therefore, $ \begin{align} 0 &=C\cdot C-2\left(\tfrac43B-\tfrac13A\right)\cdot C+\tfrac43B\cdot B-\tfrac13A\cdot A\\ &=\left|C-\left(\tfrac43B-\tfrac13A\right)\right|^2+\left(\tfrac43B\cdot B-\tfrac13A\cdot A\right)-\left(\tfrac43B-\tfrac13A\right)\cdot\left(\tfrac43B-\tfrac13A\right)\\ &=\left|C-\left(\tfrac43B-\tfrac13A\right)\right|^2-\tfrac49|A-B|^2\tag{9} \end{align} $ So the set of points that satisfy $(8)$ is the sphere centered at $\tfrac43B-\tfrac13A$ with radius $\frac23|A-B|$.