Find the vector and the Cartesian equations of the line passing through the point $(5, 2, -4)$ and parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.

(A) The position vector of the given point is $\vec{a} = 5 \hat{i} + 2 \hat{j} - 4 \hat{k}$.
The direction vector of the line is $\vec{b} = 3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.
The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.
Substituting the values,the vector equation is $\vec{r} = (5 \hat{i} + 2 \hat{j} - 4 \hat{k}) + \lambda (3 \hat{i} + 2 \hat{j} - 8 \hat{k})$.
For the Cartesian equation,let $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$.
Then $x \hat{i} + y \hat{j} + z \hat{k} = (5 + 3 \lambda) \hat{i} + (2 + 2 \lambda) \hat{j} + (-4 - 8 \lambda) \hat{k}$.
Equating the components,we get $x = 5 + 3 \lambda$,$y = 2 + 2 \lambda$,and $z = -4 - 8 \lambda$.
Solving for $\lambda$,we get $\lambda = \frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$.
Thus,the Cartesian equation is $\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$.