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The volume of the tetrahedron formed by the points $(1,1,1),(2,1,3)(3,2,2)$ and $(3,3,4)$ in cubic units is
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Verified Answer
The correct answer is:
$\frac{5}{6}$
Let $A(1,1,1), B(2,1,3), C(3,2,2)$ and $D(3,3,4)$. So, $\mathbf{A B}=\hat{\mathbf{i}}+2 \hat{\mathbf{k}}, \mathbf{A C}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{A D}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
$$
\begin{aligned}
&\text { Volume of tetrahedron }=\frac{1}{6}\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 1 \\
2 & 2 & 3
\end{array}\right] \\
&\quad=\frac{1}{6}|1(3-2)+0(6-2)+2(4-2)| \\
&=\frac{1}{6}(1+0+4)=\frac{5}{6} \text { cubic units }
\end{aligned}
$$
$$
\begin{aligned}
&\text { Volume of tetrahedron }=\frac{1}{6}\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 1 \\
2 & 2 & 3
\end{array}\right] \\
&\quad=\frac{1}{6}|1(3-2)+0(6-2)+2(4-2)| \\
&=\frac{1}{6}(1+0+4)=\frac{5}{6} \text { cubic units }
\end{aligned}
$$
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