Example 1. Find the area of the triangle whose vertices are located at points $A(1,2,3), B(3,2,1), C(1,0,1)$.
Solution. Let's find the coordinates of the vectors $\overrightarrow{A B}=\{2,0,-2\}, \overrightarrow{A C}=$ $=\{0,-2,-2\}$. We will determine the coordinates of the vector product $\overrightarrow{A B} \times$ $\times \overrightarrow{A C}($ Fig. 3.12 )

$$
\begin{aligned}
\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 0 & -2 \\
0 & -2 & -2
\end{array}\right| & =\vec{i}\left|\begin{array}{rr}
0 & -2 \\
-2 & -2
\end{array}\right|- \\
& -\vec{j}\left|\begin{array}{cc}
2 & -2 \\
0 & -2
\end{array}\right|+\vec{k}\left|\begin{array}{rr}
2 & 0 \\
0 & -2
\end{array}\right|=-4 \vec{i}+4 \vec{j}-4 \vec{k}
\end{aligned}
$$

Let's find the length of this vector, which is numerically equal to the area of the parallelogram $S$ (point $2^{\circ}$ ): $S=|\overrightarrow{A B} \times \overrightarrow{A C}|=4 \sqrt{3}$. The area of the triangle $S_{\triangle}$ is $\frac{1}{2} S$, i.e., $S_{\Delta}=\frac{1}{2} \cdot 4 \sqrt{3}=2 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-039.jpg?height=291&width=327&top_left_y=1513&top_left_x=173)

Fig. 3.12

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-039.jpg?height=411&width=471&top_left_y=1388&top_left_x=705)

Fig. 3.13