## Task Condition

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

$a=\{-7 ; 10 ;-5\}$

$b=\{0 ;-2 ;-1\}$

$c=\{-2 ; 4 ;-1\}$
## 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.

$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
-7 & 10 & -5 \\
0 & -2 & -1 \\
-2 & 4 & -1
\end{array}\right|= \\
& =-7 \cdot\left|\begin{array}{cc}
-2 & -1 \\
4 & -1
\end{array}\right|-10 \cdot\left|\begin{array}{cc}
0 & -1 \\
-2 & -1
\end{array}\right|+(-5) \cdot\left|\begin{array}{cc}
0 & -2 \\
-2 & 4
\end{array}\right|= \\
& =-7 \cdot 6-10 \cdot(-2)-5 \cdot(-4)=-42+20+20=-2
\end{aligned}
$$

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

## Problem Kuznetsov Analytic Geometry 6-30