Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).

The required line passes through the origin. Therefore, its position vector is given by,

$\vec{a}=\overrightarrow{0}$                    $\ldots(1)$

The direction ratios of the line through origin and (5, −2, 3) are

(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3

The line is parallel to the vector given by the equation, $\vec{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}$

The equation of the line in vector form through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is, $\vec{r}=\vec{a}+\lambda \vec{b}, \lambda \in R$

$\Rightarrow \vec{r}=\overrightarrow{0}+\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})$

$\Rightarrow \vec{r}=\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})$

The equation of the line through the point $\left(x_{1}, y_{1}, z_{1}\right)$ and direction ratios $a, b, c$ is given by, $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$

Therefore, the equation of the required line in the Cartesian form is

$\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}$

$\Rightarrow \frac{x}{5}=\frac{y}{-2}=\frac{z}{3}$