If the distance of point $B(2,1,-3)$ from the line passing through the point $A(4,-2,2)$, and parallel to the vector $\vec{c} = -4\hat{i} - 6\hat{j} - 2\hat{k}$ is $x$, then $x^4 + x^2 + 541 =$

The position vectors of points $A$ and $B$ are $\vec{a} = 4\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - 3\hat{k}$.

The vector joining $A$ to $B$ is:
$\overrightarrow{AB} = \vec{b} - \vec{a} = (2-4)\hat{i} + (1 - (-2))\hat{j} + (-3-2)\hat{k} = -2\hat{i} + 3\hat{j} - 5\hat{k}$

The line passes through $A$ and is parallel to the vector $\vec{c} = -4\hat{i} - 6\hat{j} - 2\hat{k}$. We can take the direction vector of the line as $\vec{d} = 2\hat{i} + 3\hat{j} + \hat{k}$.

Taking the dot product of $\overrightarrow{AB}$ and $\vec{d}$:
$\overrightarrow{AB} \cdot \vec{d} = (-2)(2) + (3)(3) + (-5)(1) = -4 + 9 - 5 = 0$

Since $\overrightarrow{AB}$ is perpendicular to the direction of the line, the perpendicular distance $x$ of point $B$ from the line is simply the magnitude of $\overrightarrow{AB}$.

$x = |\overrightarrow{AB}| = \sqrt{(-2)^2 + 3^2 + (-5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$

We need to find the value of $x^4 + x^2 + 541$.

$x^2 = 38$

$x^4 = (38)^2 = 1444$

Substituting these values:
$x^4 + x^2 + 541 = 1444 + 38 + 541 = 2023$

Answer: $2023$