The position vector of the points \(A, B, C\) are \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}),(3 \mathbf{i}-2 \mathbf{j}+\hat{\mathbf{k}})\) and \((\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})\)
respectively. These points

\(\overrightarrow{\mathbf{A B}}=(3-2) \hat{\mathbf{i}}+(-2-1) \hat{\mathbf{j}}+(1+1) \hat{\mathbf{k}}\)
\(=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \)
\( \Rightarrow|\overrightarrow{\mathbf{A B}}| =\sqrt{1+9+4}=\sqrt{14} \)
\( \overrightarrow{\mathbf{B C}} =(1-3) \hat{\mathbf{i}}+(4+2) \hat{\mathbf{j}}+(-3-1) \hat{\mathbf{k}} \)
\( =-2 \hat{\mathbf{i}}+6 \mathbf{j}-4 \hat{\mathbf{k}}\)
\(\Rightarrow \quad|\overrightarrow{\mathbf{B C}}| =\sqrt{4+36+16} \)
\( =\sqrt{56}=2 \sqrt{14} \)
\( \overrightarrow{\mathbf{C A}} =(2-1) \hat{\mathbf{i}}+(1-4) \hat{\mathbf{j}}+(-1+3) \hat{\mathbf{k}} \)
\( =\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \)
\( \Rightarrow \quad|\overrightarrow{\mathbf{C A}}| =\sqrt{1+9+4}=\sqrt{14}\)
So, \(|\overrightarrow{\mathbf{A B}}|+|\overrightarrow{\mathbf{A C}}|=|\overrightarrow{\mathbf{B C}}|\) and angle between
\(A B\) and \(B C\) is \(180^{\circ} .\)
\(\therefore\) Points \(A, B, C\) cannot form an isosceles triangle. Hence, \(A, B, C\) are collinear.