Find the vector and the cartesian equations of the line that passes through the origin and $(5, - 2,3)$.

Let $\vec a$ and $\vec b$ be the position vectors of point $A(0,0,0)$ and $B(5, - 2,3)$

Let $\vec a = 0\hat i + 0\hat j + 0\hat k$ and $\vec b = 5\hat i - 2\hat j + 3\hat k$
$\Rightarrow$ $\vec b - \vec a = 5\hat i - 2\hat j + 3\hat k$

Let $\vec r$ be the position vector of any point on the line. Then, the vector equation of line is

$\vec r = \vec a + \lambda (\vec b - \vec a) = \hat 0 + \lambda (5\hat i - 2\hat j + 3\hat k)$

Now, $\vec r = \vec a + \lambda (\vec b - \vec a) = \vec 0 + \lambda (5\hat i - 2\hat j + 3\hat k)$

Eliminating $\lambda$,

we get $\cfrac{x}{5} = - \cfrac{y}{2} = \cfrac{z}{3}$ is the equation of the line in cartesian form.