Find the vector and cartesian equation of the line through the point $\left(3, -4, -2\right)$ and parallel to the vector $9 \hat{i} + 6 \hat{j} + 2 \hat{k}$.
Let $\overrightarrow{a}$ be the position vector of the point $\left(3, -4, -2\right)$.
Then, $\;$ $\overrightarrow{a} = 3 \hat{i} - 4 \hat{j} - 2 \hat{k}$
Let $\;$ $\overrightarrow{m} = 9 \hat{i} + 6 \hat{j} + 2 \hat{k}$
Then, the vector equation of the required line is $\;\;$ $\overrightarrow{r} = a + \lambda \overrightarrow{m}$ $\;\;$ where $\lambda$ is a constant.
i.e. $\;$ $\overrightarrow{r} = \left(3 \hat{i} - 4 \hat{j} - 2 \hat{k}\right) + \lambda \left(9 \hat{i} + 6 \hat{j} + 2 \hat{k}\right)$ $\;\;\; \cdots \; (1)$
Equation $(1)$ is the required vector equation of the line.
Now, $\overrightarrow{r}$ is the position vector of any point $P \left(x, y, z\right)$ on the line.
i.e. $\;$ $\overrightarrow{r} = x \hat{i} + y \hat{j} + z \hat{k}$
$\begin{aligned}
\therefore \; x \hat{i} + y \hat{j} + z \hat{k} & = \left(3 \hat{i} - 4 \hat{j} - 2 \hat{k}\right) + \lambda \left(9 \hat{i} + 6 \hat{j} + 2 \hat{k}\right) \\\\
& = \left(3 + 9 \lambda\right) \hat{i} + \left(-4 + 6 \lambda\right) \hat{j} + \left(-2 + 2 \lambda\right) \hat{k}
\end{aligned}$
Comparing the coefficients of $\hat{i}$, $\hat{j}$ and $\hat{k}$ on both sides, we get
$x = 3 + 9 \lambda$, $\;\;$ $y = -4 + 6 \lambda$ $\;\;$ $z = -2 + 2 \lambda$
Eliminating $\lambda$,
$\dfrac{x - 3}{9} = \dfrac{y + 4}{6} = \dfrac{z + 2}{2}$ $\;\;\; \cdots \; (2)$
Equation $(2)$ gives the equation of the line in cartesian form.