Using the section formula,show that the points $A(2, -3, 4)$,$B(-1, 2, 1)$,and $C(0, \frac{1}{3}, 2)$ are collinear.

(N/A) Let the point $C$ divide the line segment $AB$ in the ratio $k:1$.
By the section formula,the coordinates of the point dividing $AB$ are $\left(\frac{k(-1) + 2}{k+1}, \frac{k(2) - 3}{k+1}, \frac{k(1) + 4}{k+1}\right)$.
Equating these to the coordinates of point $C(0, \frac{1}{3}, 2)$,we have:
$\frac{-k+2}{k+1} = 0 \implies -k+2 = 0 \implies k = 2$.
Now,check the other coordinates for $k=2$:
$y$-coordinate: $\frac{2(2)-3}{2+1} = \frac{4-3}{3} = \frac{1}{3}$.
$z$-coordinate: $\frac{2(1)+4}{2+1} = \frac{6}{3} = 2$.
Since the coordinates match for $k=2$,point $C$ lies on the line segment $AB$ dividing it in the ratio $2:1$.
Therefore,the points $A, B,$ and $C$ are collinear.