Let $\vec{a}$ and $\vec{b}$ be the position vectors of point A = (3, –2, –5) and B = (3, –2, 6). Let $\vec{a}=3\hat{i}-2\hat{j}-5\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+6\hat{k}$ $\Rightarrow \vec{b}-\vec{a}=0\hat{i}+0\hat{j}+11\hat{k}$ 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{r}=3\hat{i}-2\hat{j}-5\hat{k}+\lambda(11\hat{k})$Now, $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ $=3\hat{i}-2\hat{j}+(-5+\lambda)\hat{k}$Eliminating λ, we get $\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}$ is the equation of line in cartesian form.