Let the points A(2, –3, 4), B(–1, 2, 1) and C $\left(0, \frac{1}{3}, 2\right)$ be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of P are $\left( \frac{-k+2}{k+1}, \frac{2k-3}{k+1}, \frac{k+4}{k+1} \right)$ Let us examine whether for some value of k, the point P coincides with point C. On putting $\frac{-k+2}{k+1}$ = 0 , we get k = 2 When k = 2 , then $\frac{2k-3}{k+1}$ = $\frac{4-3}{3}$ = $\frac{1}{3}$ and $\frac{k+4}{k+1}$ = $\frac{2+4}{2+1}$ = 2 Therefore, C $\left(0, \frac{1}{3}, 2\right)$ is a point which divides AB internally in the ratio 2 : 1. Hence A, B, C are collinear.