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The volume of the tetrahedron having the edges $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ as coterminous, is $\frac{2}{3}$ cubic unit. Then $\lambda$ equals
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$\begin{aligned}
& \text { Let } \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& \text { and } \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}
\end{aligned}$
Since, volume of tetrahedron $=\frac{1}{6}[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
$\begin{array}{ll}
\Rightarrow & \frac{2}{3}=\frac{1}{6}\left|\begin{array}{rrr}
1 & 2 & -1 \\
1 & 1 & 1 \\
1 & -1 & \lambda
\end{array}\right| \\
\Rightarrow & \frac{2}{3}=\frac{1}{6}[1(\lambda+1)-2(\lambda-1)-1(-1-1)] \\
\Rightarrow & 4=[-\lambda+5] \Rightarrow \lambda=1
\end{array}$
& \text { Let } \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& \text { and } \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}
\end{aligned}$
Since, volume of tetrahedron $=\frac{1}{6}[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
$\begin{array}{ll}
\Rightarrow & \frac{2}{3}=\frac{1}{6}\left|\begin{array}{rrr}
1 & 2 & -1 \\
1 & 1 & 1 \\
1 & -1 & \lambda
\end{array}\right| \\
\Rightarrow & \frac{2}{3}=\frac{1}{6}[1(\lambda+1)-2(\lambda-1)-1(-1-1)] \\
\Rightarrow & 4=[-\lambda+5] \Rightarrow \lambda=1
\end{array}$
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