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The volume (in cubic units) of the tetrahedron with edges $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ is
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$2 / 3$
We know that, volume of tetrahedron
$\begin{aligned} & =\frac{1}{6}[\overrightarrow{\mathbf{A B}} \overrightarrow{\mathbf{A C}} \overrightarrow{\mathbf{A D}}] \\ & =\frac{1}{6}\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1\end{array}\right|\end{aligned}$
$\begin{aligned} & =\frac{1}{6}[1(1-2)-1(-1-1)+1(2+1)] \\ & =\frac{1}{6}[-1+2+3] \\ & =\frac{4}{6}=\frac{2}{3}\end{aligned}$
$\begin{aligned} & =\frac{1}{6}[\overrightarrow{\mathbf{A B}} \overrightarrow{\mathbf{A C}} \overrightarrow{\mathbf{A D}}] \\ & =\frac{1}{6}\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1\end{array}\right|\end{aligned}$
$\begin{aligned} & =\frac{1}{6}[1(1-2)-1(-1-1)+1(2+1)] \\ & =\frac{1}{6}[-1+2+3] \\ & =\frac{4}{6}=\frac{2}{3}\end{aligned}$
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