2.13. Are the vectors coplanar

$$
\vec{a}, \vec{b}, \vec{c}: \vec{a}=\{1,-2,1\}, \vec{b}=\{3,1,-2\}, \vec{c}=\{7,14,-13\} ?
$$
Solution. To answer the given question, it is necessary to calculate the mixed product of these vectors, and if it turns out to be zero, this will indicate that the vectors $\ddot{a}, \vec{b}, \vec{c}$ are coplanar. We find the mixed product $\vec{a} \cdot \vec{b} \cdot \vec{c}$ using formula (2.40).

$$
\begin{aligned}
\vec{a} \times \vec{b} \cdot \vec{c} & =\left|\begin{array}{ccc}
1 & -2 & 1 \\
3 & 1 & -2 \\
7 & 14 & -13
\end{array}\right|=1\left|\begin{array}{cc}
1 & -2 \\
14 & -13
\end{array}\right|-(-2)\left|\begin{array}{cc}
3 & -2 \\
7 & -13
\end{array}\right|+1\left|\begin{array}{cc}
3 & 1 \\
7 & 14
\end{array}\right|= \\
& =-13+28+2(-39+14)+42-7=15-50+35=0
\end{aligned}
$$

Since the mixed product $\vec{a} \cdot \vec{b} \cdot \vec{c}$ turned out to be zero, the vectors $\vec{a}, \ddot{b}, \vec{c}$ are coplanar.