Show that the points $A(-2\hat{i}+3\hat{j}+5\hat{k}), B(\hat{i}+2\hat{j}+3\hat{k})$ and $C(7\hat{i}-\hat{k})$ are collinear

Question

Show that the points $A(-2\hat{i}+3\hat{j}+5\hat{k}), B(\hat{i}+2\hat{j}+3\hat{k})$ and $C(7\hat{i}-\hat{k})$ are collinear.

Can I prove it using the section formula of vectors ?

My Attempt $$ \overrightarrow{OC}=\frac{\overrightarrow{OA}+n.\overrightarrow{OB}}{1+n}\implies 7\hat{i}-\hat{k}=\frac{-2\hat{i}+3\hat{j}+5\hat{k}+n(\hat{i}+2\hat{j}+3\hat{k})}{1+n}\\ 7\hat{i}-\hat{k}+7n\hat{i}-n\hat{k}=-2\hat{i}+3\hat{j}+5\hat{k}+n\hat{i}+2n\hat{j}+3n\hat{k}\\ (9+6n)\hat{i}+(-3-2n)\hat{j}+(-6-4n)\hat{k}=0\\ 9+6n=0\implies n=3/2\\ -3-2n=0\implies n=3/2\\ -6-4n=0\implies n=3/2 $$ Thus C divides AB in a ratio $2:3$, satisfies the section formula. ie., A,B,C are collinear

Answer 1

Yes you can.

Also, we can use the following. $$\vec{AB}=3\vec{i}-\vec{j}-2\vec{k}$$ and $$\vec{AC}=9\vec{i}-3\vec{j}-6\vec{k}=3\vec{AB},$$ which says $AB||AC$ and these points are collinear.