The position vectors of the points A, B, C are $2\hat i+\hat j-\hat k, 3\hat i-2\hat j +\hat k$ and $\hat i +4\hat j - 3\hat k$ respectively. These points

are collinear

Let $\vec a=2\hat i+\hat j-\hat k, \vec b=3\hat i-2\hat j+\hat k$ and $\vec c=\hat i+4\hat j-3\hat k$. Then,

$\vec{AB}=\vec b-\vec a=\hat i-3\hat j+2\hat k$ and, $\vec{BC}=-2\hat i +6\hat j-4\hat k$

Clearly, $\vec{BC}=-2\vec{AB}=2\vec{BA}$

So, points A, B, C are collinear.