The three vectors  $\stackrel{\wedge }{i}+\stackrel{\wedge }{j},\stackrel{\wedge }{j}+\stackrel{\wedge }{k},\stackrel{\wedge }{k}+\stackrel{\wedge }{i}$ taken two at a time form three planes, If V be the volume of the tetrahedron having adjacent sides as the three unit vectors drawn perpendicular to those three planes then the value of $9\sqrt{3}V$ is 

$$\begin{gathered} Normal vector of the first plane is given by \\ (\overline{i}+\overline{j}) \times (\overline{j}+\overline{k})=\overline{i}-\overline{j}+\overline{k} \\ It\text{'}s unit vector is \frac{\overline{i}-\overline{j}+\overline{k}}{\sqrt{3}}={n^{^}}_1 \\ The normal vector of second plane is given by \\ (\overline{j}+\overline{k})\times (\overline{k}+\overline{i}) =\overline{i}+\overline{j}-\overline{k } \\ It\text{'}s unit vector is \frac{\overline{i}+\overline{j}-\overline{k }}{\sqrt{3}}={n^{^}}_2 \\ The normal vector of third plane is given by \\ (\overline{k}+\overline{i}) \times (\overline{i}+\overline{j}) =-\overline{i}+\overline{j}+\overline{k} \\ It\text{'}s unit vector is \frac{-\overline{i}+\overline{j}+\overline{k}}{\sqrt{3}}={{n^{^}}_3}^{ } \\ Now volume of tetrahedron =\frac{1}{6}\left[{n^{^}}_1 {n^{^}}_2 {n^{^}}_3\right] \end{gathered}$$

$$\begin{gathered} V=\frac{1}{6}\times \frac{1}{3\sqrt{3}}\left|\begin{array}{ccc}1& -1& 1\\ 1& 1& -1\\ -1& 1& 1\end{array}\right| \\ =\frac{2}{9\sqrt{3}} \end{gathered}$$

$\Rightarrow 9\sqrt{3}V=2$