Show that the vectors $\hat{i} - 2 \hat{j} + 3 \hat{k}$, $-2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $-\hat{j} + 2 \hat{k}$ are coplanar.
Let $\alpha$, $\beta$ and $\gamma$ be three constants.
Let $\overrightarrow{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$;
$\overrightarrow{b} = -2 \hat{i} + 3 \hat{j} - 4 \hat{k}$;
$\overrightarrow{c} = - \hat{j} + 2 \hat{k}$
Three vectors are coplanar if $\alpha \overrightarrow{a} + \beta \overrightarrow{b} + \gamma \overrightarrow{c} = \overrightarrow{0}$
i.e. if $\alpha \left(\hat{i} - 2 \hat{j} + 3 \hat{k}\right) + \beta \left(-2 \hat{i} + 3 \hat{j} - 4 \hat{k}\right) + \gamma \left(- \hat{j} + 2 \hat{k}\right) = \overrightarrow{0}$ $\;\;\; \cdots \; (1)$
i.e. if
$\hat{i}$ terms: $\;$ $\alpha - 2 \beta + 0 \gamma = 0$ $\;\;\; \cdots \; (2a)$
$\hat{j}$ terms: $\;$ $- 2 \alpha + 3 \beta - \gamma = 0$ $\;\;\; \cdots \; (2b)$
$\hat{k}$ terms: $\;$ $3 \alpha - 4 \beta + 2 \gamma = 0$ $\;\;\; \cdots \; (2c)$
Solving equations $(2a)$, $(2b)$ and $(2c)$ using the Gauss method, the augmented matrix is
$\begin{bmatrix}
1 & -2 & 0 & | 0 \\
-2 & 3 & -1 & | 0 \\
3 & -4 & 2 & | 0
\end{bmatrix}$
$R_3 - 3 R_1$ $\implies$
$\begin{bmatrix}
1 & -2 & 0 & | 0 \\
-2 & 3 & -1 & | 0 \\
0 & 2 & 2 & | 0
\end{bmatrix}$
$R_2 + 2R_1$ $\implies$
$\begin{bmatrix}
1 & -2 & 0 & | 0 \\
0 & -1 & -1 & | 0 \\
0 & 2 & 2 & | 0
\end{bmatrix}$
$R_3 + 2 R_2$ $\implies$
$\begin{bmatrix}
1 & -2 & 0 & | 0 \\
0 & -1 & -1 & | 0 \\
0 & 0 & 0 & | 0
\end{bmatrix}$
$\implies$ The system has many solutions.
$\therefore$ $\;$ $\exists$ non zero combination of numbers $\alpha$, $\beta$, $\gamma$ such that the linear combination of $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ is equal to $\overrightarrow{0}$.
For example: $\;$ $2 \overrightarrow{a} + \overrightarrow{b} - \overrightarrow{c} = \overrightarrow{0}$
$\implies$ Vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are linearly dependent.
$\implies$ $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are coplanar.