A vector  $\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}+\beta \widehat{k}, \left(\alpha ,\beta \in R\right)$ lies in the plane of the vectors $\overrightarrow{b}=\widehat{i}+\widehat{j}$ and $\overrightarrow{c}=\widehat{i}-\widehat{j}+4\widehat{k}$. If $\overrightarrow{a}$bisects the angle between $\overrightarrow{b}$ and$\overrightarrow{c}$, . then

Angle bisector can be  $\overrightarrow{a}=\lambda \left(\widehat{b}+\widehat{c}\right) or \overrightarrow{a}=\mu \left(\widehat{b}-\widehat{c}\right)$

Now $\overrightarrow{a}=\lambda \left(\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}+\frac{\widehat{i}-\widehat{j}+4\widehat{k}}{3\sqrt{2}}\right)=\frac{\lambda }{3\sqrt{2}}\left[3\widehat{i}+3\widehat{j}+\widehat{i}-\widehat{j}+4\widehat{k}\right]=\frac{\lambda }{3\sqrt{2}}\left[4\widehat{i}+2\widehat{j}+4\widehat{k}\right]$          

Comparing  with $\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}+\beta \widehat{k}$ we get

                                               $\frac{2\lambda }{3\sqrt{2}}=2\Rightarrow \lambda =3\sqrt{2}$         

 $\overrightarrow{\Rightarrow a}=4\widehat{i}+2\widehat{j}+4\widehat{k}$         

          None of the option is satisfied. So, now consider

$\overrightarrow{a}=\mu \left(\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}-\frac{\widehat{i}-\widehat{j}+4\widehat{k}}{3\sqrt{2}}\right)$

      $\Rightarrow \overrightarrow{a}=\frac{\mu }{3\sqrt{2}}\left(3\widehat{i}+3\widehat{j}-\widehat{i}+\widehat{j}-4\widehat{k}\right)$

               $=\frac{\mu }{3\sqrt{2}}\left(2\widehat{i}+4\widehat{j}-4\widehat{k}\right)$                     

          Comparing with$\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}+\beta \widehat{k}$, we get

    $\frac{4\mu }{3\sqrt{2}}=2\Rightarrow \mu =\frac{3\sqrt{2}}{2}$

$\Rightarrow \overrightarrow{a}=\widehat{i}+2\widehat{j}-2\widehat{k}$

Now $\overrightarrow{a}.\widehat{k}+2=-2+2=0$