Let the points be A, B and C with position vectors $3\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}},\ \hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and $-\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$ respectively. Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A $=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}-3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
$=-2\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$
Position vector of C - Position vector of B $=-\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$=-2\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, A, B, and C are collinear.