• # question_answer The position vectors of three points A, B, C are $\hat{i}+2\hat{j}+3\hat{k}\cdot 2\hat{i}+3\hat{j}+\hat{k}\And 3\hat{i}+\hat{j}+2\hat{k}.$ A unit vector perpendicular to the plane of the triangle ABC is: A) $\left( -\frac{1}{\sqrt{3}} \right)\,\,(\hat{i}+\hat{j}+\hat{k})$ B) $\left( \frac{1}{\sqrt{3}} \right)\,\,(\hat{i}-\hat{j}+\hat{k})$ C) $\left( \frac{1}{\sqrt{3}} \right)\,\,(\hat{i}+\hat{j}-\hat{k})$ D) none of these

 unit vector $\bot$ to plane ABC $=\frac{\overrightarrow{AB}\times \overrightarrow{AC}}{|\overrightarrow{AB}\times \overrightarrow{AC}|}$$=\frac{1}{|\overrightarrow{AB}\times \overrightarrow{AC}|}\,\,\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 1 & -\,2 \\ 2 & -1 & -1 \\ \end{matrix} \right|$$=\frac{\hat{i}\,(-1-2)-\hat{j}\,(-1+4)+\hat{k}\,(-1-2)}{|\overrightarrow{AB}\times \overrightarrow{AC}|}$$=\frac{3\hat{i}-3\hat{j}-3\hat{k}}{\sqrt{27}}=\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}$