Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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