The distance of the point $(5,3,-1)$ from the plane passing through points $(2,1,0),(3,-2,4)$ and $(1,-3,3)$…

The plane containing points $A=(2,1,0)$, $B=(3,-2,4)$, and $C=(1,-3,3)$ has the vectors $\vec{AB} = (1, -3, 4)$ and $\vec{AC} = (-1, -4, 3)$.

A normal vector to the plane is given by their cross product:
$\vec{n} = \vec{AB} \times \vec{AC} = 7\mathbf{i} - 7\mathbf{j} - 7\mathbf{k}$.
This simplifies to $\vec{n} = (1, -1, -1)$ by scaling.

Using the point-normal form with $A=(2,1,0)$ and normal $(1,-1,-1)$:
$1(x-2) - 1(y-1) - 1(z-0) = 0$
Simplifying yields the plane equation $x - y - z - 1 = 0$.

The distance from point $(5,3,-1)$ to this plane is calculated as:

$D = \frac{|1(5) + (-1)(3) + (-1)(-1) + (-1)|}{\sqrt{1^2 + (-1)^2 + (-1)^2}} = \frac{|2|}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

This corresponds to option A.