Solution:

Given:
The origin $(0,0,0)$ and the point $(5,-2,3)$
It is known to us that
The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
Here, the position vectors of the two points $(0,0,0)$ and $(5,-2,3)$ are $\vec{a}=0 \hat{i}+0 \hat{j}+0 \hat{k}$ and $\vec{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}$, respectively.
Therefore, the vector eq. of the required line is given as:
$\begin{array}{l}
\overrightarrow{\mathrm{r}}=0 \hat{\mathrm{i}}+0 \hat{\mathbf{j}}+0 \hat{\mathrm{k}}+\lambda[(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})-(0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}})] \\
\overrightarrow{\mathrm{r}}=\lambda(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})
\end{array}$
By using the formula,now
The Cartesian eq. of a line that passes through two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\left(\mathrm{x}_{2}\right.$, $\left.y_{2}, z_{2}\right)$ is given as $\frac{x-X_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$

Therefore, the Cartesian equation of the line that passes through the origin $(0,0,0)$ and $(5,-2,3)$ is
$\frac{x-0}{5-0}=\frac{y-0}{-2-0}=\frac{z-0}{3-0} \Rightarrow \frac{x}{5}=\frac{y}{-2}=\frac{z}{3}$
As a result, the vector equation is
$\overrightarrow{\mathrm{r}}=\lambda(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
The Cartesian equation is
$\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}$