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

(A) The required line passes through the origin $(0, 0, 0)$ and the point $(5, -2, 3).$
The position vector of the origin is $\vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}.$
The direction ratios of the line passing through $(0, 0, 0)$ and $(5, -2, 3)$ are $(5-0, -2-0, 3-0) = (5, -2, 3).$
Thus,the line is parallel to the vector $\vec{b} = 5\hat{i} - 2\hat{j} + 3\hat{k}.$
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r} = \vec{a} + \lambda\vec{b}.$
Substituting the values,we get $\vec{r} = \vec{0} + \lambda(5\hat{i} - 2\hat{j} + 3\hat{k}) = \lambda(5\hat{i} - 2\hat{j} + 3\hat{k}),$ where $\lambda \in R.$
The Cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$
Substituting $(x_1, y_1, z_1) = (0, 0, 0)$ and $(a, b, c) = (5, -2, 3),$ we get $\frac{x-0}{5} = \frac{y-0}{-2} = \frac{z-0}{3}.$
Therefore,the Cartesian equation is $\frac{x}{5} = \frac{y}{-2} = \frac{z}{3}.$