Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1)

The given points are $\mathrm{A}(2,-3,4), \mathrm{B}(-1,2,1)$, and $\mathrm{C}\left(0, \frac{1}{3}, 2\right)$

Let P be a point that divides AB in the ratio k:1.

Hence, by section formula, the coordinates of P are given by

$\left(\frac{k(-1)+2}{k+1}, \frac{k(2)-3}{k+1}, \frac{k(1)+4}{k+1}\right)$

Now, we find the value of k at which point P coincides with point C.

By taking $\frac{-k+2}{k+1}=0$, we obtain $k=2$.

For $k=2$, the coordinates of point $\mathrm{P}$ are $\left(0, \frac{1}{3}, 2\right)$.

i.e., $\mathrm{C}\left(0, \frac{1}{3}, 2\right)$ is a point that divides AB externally in the ratio $2.1$ and is the same as point P.

Hence, points A, B, and C are collinear.