## Condition of the problem

Are the vectors $a, b$ and $c$ coplanar?

$$
\begin{aligned}
& a=\{2 ; 3 ; 1\} \\
& b=\{-1 ; 0 ;-1\} \\
& c=\{2 ; 2 ; 2\}
\end{aligned}
$$
## Solution

For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.

$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
2 & 3 & 1 \\
-1 & 0 & -1 \\
2 & 2 & 2
\end{array}\right|= \\
& =2 \cdot\left|\begin{array}{cc}
0 & -1 \\
2 & 2
\end{array}\right|-3 \cdot\left|\begin{array}{cc}
-1 & -1 \\
2 & 2
\end{array}\right|+1 \cdot\left|\begin{array}{cc}
-1 & 0 \\
2 & 2
\end{array}\right|= \\
& =2 \cdot 2-3 \cdot 0+1 \cdot(-2)=4-0-2=2
\end{aligned}
$$

Since $(a, b, c)=2 \neq 0$, the vectors $a, b$, and $c$ are not coplanar.

## Problem Kuznetsov Analytic Geometry 6-1