Let the points A(2, –3, 4), B(–1, 2, 1) and C (0,

1

3

,2) be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of P are (

−k+2

k+1

,

2k−3

k+1

,

k+4

k+1

)
Let us examine whether for some value of k, the point P coincides with point C.
On putting

−k+2

k+1

= 0 , we get k = 2
When k = 2 , then

2k−3

k+1

=

4−3

3

=

1

3

and

k+4

k+1

=

2+4

2+1

= 2
Therefore, C (0,

1

3

,2) is a point which divides AB internally in the ratio 2 : 1. Hence A, B, C are collinear.