This way and that

Let the respective lengths of the six successive steps be $a,b,c,d,e,f.$ Then we have the condition that

$a + b + c + d + e + f = 10$ such that each variable is an integer $\ge 1.$

We can simplify this condition by replacing each variable $x$ with $x'$ such that $x' = x - 1.$ Then the equation become

$a' + b' + c' + d' + e' + f' = 4$ such that each variable is an integer $\ge 0.$

This is a 'stars and bars' equation with $\binom{9}{4} = 126$ solutions.

Now what matters here as far as calculating $S$ is concerned is the total distances east, north and up the particle moves after six steps. To this end, we focus on the (ordered) partitions of $4$ into three non-negative integers. These partitions are $(4,0,0), (3,1,0), (2,2,0), (2,1,1).$ For each of these partitions, we can assign one element to $A = (a + d),$ one to $B = (b + e)$ and one to $C = (c + f).$ That is, we could assign the ordered triplet $(A,B,C)$ any of the permutations of these four partitions. So we now need to look at each of these partitions separately.

• $(4,0,0)$ : There are $3$ triplets $(A,B,C)$ associated with this partition. With, for example, $A = 4,$ we must then look at the equation $a + d = 4$ where $a,d,$ are non-negative integers. This is again a stars and bars equation with $\binom{5}{4} = 5$ solutions. Then with $B = b + e = 0$ and $C = c + f = 0$ each being "solved" in one way only, we have $5*1*1 = 5$ paths associated with the triplet $(A,B,C) = (4,0,0),$ and since there are $3$ permutations of this partition we have $3*5 = 15$ paths associated with this partition. For each of these paths, the distance between the starting and finishing points of the particle is $\sqrt{6^{2} + 2^{2} + 2^{2}} = \sqrt{44} = 2\sqrt{11}.$

• $(3,1,0)$ : There are $6$ triplets associated with this partition. With, for example, $A = 3, B = 1$ and $C = 0,$ we can solve $A = a + d = 3$ in $4$ ways, $B = b + e = 1$ in $2$ ways and $C = c + f = 0$ in one way. Thus there are $6*4*2*1 = 48$ paths associated with this partition, each having a "distance" of $\sqrt{5^{2} + 3^{2} + 2^{2}} = \sqrt{38}.$

• $(2,2,0)$ : Quickly, we have $3$ triplets, each associated with $3*3*1 = 9$ paths for a total of $27$ paths, each with a distance of $\sqrt{4^{2} + 4^{2} + 2^{2}} = 6.$

• $(2,1,1)$ : Again quickly, we have $3$ triplets, each associated with $3*2*2 = 12$ paths for a total of $36$ paths, each with a distance of $\sqrt{4^{2} + 3^{2} + 3^{2}} = \sqrt{34}.$

Thus $S = \dfrac{1}{126}(15*2\sqrt{11} + 48*\sqrt{38} + 27*6 + 36*\sqrt{34}) = 6.089721.....,$

and so $\lfloor 1000*S \rfloor = \boxed{6089}.$