Find the volume of tetrahedron having edges $\overline{i}+\overline{j}+\overline{k}, \overline{i}-\overline{j}, \overline{ i}+2\overline{j}+\overline{k}$

Let $\overline{a}=\overline{i}+\overline{j}+\overline{k},\overline{b}=\overline{i}-\overline{j}$ and $\overline{c}=\overline{i}+2\overline{j}+\overline{k}$

$\therefore$ Volume of the tetrahedron $=\frac{1}{6}\left|\left[\begin{array}{ccc}\overline{a}& \overline{b}& \overline{c}\end{array}\right]\right|$

$=\frac{1}{6}\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 0\\ 1& 2& 1\end{array}\right|$

$=\frac{1}{6}\left[1\right(-1-0)-1(1-0)+1(2+1\left)\right]$

$=\frac{1}{6}[-1-1+3]=\frac{1}{6}\left[1\right]=\frac{1}{6}$ cubic units