If $\overrightarrow{A}=2\hat{i}+3\hat{j}+6\hat{k}$  and$\overrightarrow{B}=3\hat{i}-6\hat{j}+2\hat{k}$ , then vector perpendicular to both $\overrightarrow{A}$  and $\overrightarrow{B}$  has magnitude k times that of$\left(6\hat{i}+2\hat{j}-3\hat{k}\right)$ . Then k is equal to

Let $\overrightarrow{C}$  be a vector perpendicular to $\overrightarrow{A}$  and $\overrightarrow{B}$ 

Then as per question $k\overrightarrow{C}=\overrightarrow{A}\times \overrightarrow{B}$

or   $k=\frac{\left(\overrightarrow{A}\times \overrightarrow{B}\right)}{\overrightarrow{C}}$

          $=\frac{\left(2\hat{i}+3\hat{j}+6\hat{k}\right)\times \left(3\hat{i}-6\hat{j}+2\hat{k}\right)}{\left(6\hat{i}+2\hat{j}-3\hat{k}\right)}$

         $=\frac{(42\hat{i}+14\hat{j}-21\hat{k})}{(6\hat{i}+2\hat{j}-3\hat{k})}=7$$$\begin{gathered} \left(\because \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0 \\ \hat{i}\times \hat{j}=\hat{k} \hat{j}\times \hat{k}=\hat{i} \hat{k}\times \hat{i}=\hat{j}\right) \end{gathered}$$