Given points are $A(2,-3,4),\:B(-1,2,1),\: C(0,\large\frac{1}{3}$$,2)$

If the three points are collinear then according to section formula,

$C$ divides $AB$ in in some ratio.

Let the ratio in which $C$ divides $AB$ be $k:1$

We know that from section formula,

the coordinates of $C$ is given by $C\big(\large\frac{-k+2}{k+1},\frac{2k-3}{k+1},\frac{k+4}{k+1}\big)$

$\therefore$ By comparing the given coordinates of $C(0,\large\frac{1}{3}$$,2)$

we get $\large\frac{-k+2}{k+1}$$=0$, $\large\frac{2k-3}{k+1}=\frac{1}{3}$ and $\large\frac{k+4}{k+1}$$=2$

$\Rightarrow\:k=2$ in all the three equations.

$\therefore$ The given three points are collinear.