Find the vector and the cartesian equations of the line that passes through the points $(3,-2,-5)$ and $(3,-2,6).$

(D) Let the line passing through the points $P(3,-2,-5)$ and $Q(3,-2,6)$ be $PQ.$
Since the line passes through $P(3,-2,-5),$ its position vector $\vec{a}$ is given by $\vec{a} = 3\hat{i} - 2\hat{j} - 5\hat{k}.$
The direction ratios of the line $PQ$ are $(3-3, -2-(-2), 6-(-5)) = (0, 0, 11).$
The vector $\vec{b}$ in the direction of the line is $\vec{b} = 0\hat{i} + 0\hat{j} + 11\hat{k} = 11\hat{k}.$
The vector equation of the line is $\vec{r} = \vec{a} + \lambda\vec{b},$ where $\lambda \in R.$
Substituting the values,we get $\vec{r} = (3\hat{i} - 2\hat{j} - 5\hat{k}) + \lambda(11\hat{k}).$
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 the points and direction ratios,we get $\frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{11}.$