Let $\vec{a}$ and $\vec{b}$ and 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}$$\Rightarrow\vec{b}-\vec{a}=5\hat{i}-2\hat{j}+3\hat{k}$Let $\vec{r}$ be the position vector of any point on the line. Then the vector equation of line is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$ $=\vec{0}+\lambda(5\hat{i}-2\hat{j}+3\hat{k})$Now, $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\lambda(5\hat{i}-2\hat{j}+3\hat{k})$Eliminating λ, we get $\frac{x}{5}=-\frac{y}{2}=\frac{z}{3}$ is the equation of the line in cartesian form.