Find the vector and cartesian equations of the plane which passes through the point $(5, 2, -4)$ and is perpendicular to the line with direction ratios $2, 3, -1$.

(D) The position vector of the point $(5, 2, -4)$ is $\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}$.
Since the plane is perpendicular to the line with direction ratios $2, 3, -1$,the normal vector to the plane is $\vec{N} = 2\hat{i} + 3\hat{j} - \hat{k}$.
The vector equation of a plane passing through point $\vec{a}$ and perpendicular to normal $\vec{N}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$.
Substituting the values,we get $(\vec{r} - (5\hat{i} + 2\hat{j} - 4\hat{k})) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$.
To find the Cartesian equation,let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
Then $((x - 5)\hat{i} + (y - 2)\hat{j} + (z + 4)\hat{k}) \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$.
This simplifies to $2(x - 5) + 3(y - 2) - 1(z + 4) = 0$.
Expanding this,we get $2x - 10 + 3y - 6 - z - 4 = 0$,which results in $2x + 3y - z = 20$.