Solution:

Given:
Vector equation of a line that passes through a given point whose position vector is $\vec{a}$ and parallel to a given vector $\vec{b}$ is $\vec{r}=\vec{a}+\lambda \vec{b}$
Let, $\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$
Therefore, the vector equation of the required line is:
$\overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})$
The Cartesian eq. of a line through a point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and having direction cosines $1, \mathrm{~m}, \mathrm{n}$ is given by
$\frac{\mathrm{x}-\mathrm{x}_{1}}{1}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~m}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{n}}$
Here, $x_{1}=2, y_{1}=-1, z_{1}=4$ and $a=1, b=2, c=-1$
$\therefore$ The Cartesian equation of the required line is:
$\frac{x-2}{1}=\frac{y-(-1)}{2}=\frac{z-4}{-1} \Rightarrow \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}$