If the points $(2,-3,4)$, $(-1,2,1)$ and $(k,1/3,2)$ are collinear, then find the value of $k$.

If three points $A,B,C$ are collinear, the direction ratios of $AB$ and $BC$ must be proportional.Direction ratios of $AB$: $(-1-2,2-(-3),1-4)=(-3,5,-3)$.Direction ratios of $BC$: $(k-(-1),1/3-2,2-1)=(k+1,-5/3,1)$.Since they are proportional:
$$\frac{k+1}{-3}=\frac{-5/3}{5}=\frac{1}{-3}$$
From $\frac{-5/3}{5}=\frac{1}{-3}$:
$$-1/3=-1/3$$
(This confirms the ratio is $-1/3$).Now solve for $k$:
$$\frac{k+1}{-3}=-\frac{1}{3}$$
$$k+1=1⟹k=0$$