I start by restating and modifying the problem slightly, then solve it under this new
form; the passage to the initial problem is canonical.
Problem.
Consider the set
$S=S(d+4)=\{x_1, x_2, x_3, y_1, y_2, y_3, z_1, \ldots, z_{d-2}\}$. A
permutation of $S$ is called $x,y$-adjacent free (and later of
type $(1,1)$) if no two $x$'s and no two $y$'s are adjacent.
Determine the number of $x,y$-adjacent free permutations.
Note that the elements $x_1$, $x_2$, and $x_3$ are seen as distinct
elements (and the same for the $y$'s).
The passage from Problem to the initial question is given by the obvious
formula
$$
\frac{\text{number of $x,y$-adjacent free permutations of $S$}}
{3!\, 3!\, (d-2)!}
$$
in which we take $d=3$.
The solution leans towards recursive algebraic computations (the model
being Pascal's triangle). On the positive side, we might be happy with recursive
formulas (generalizations, computations of other similar numbers...). On the negative side, there will be cumbersome flying indices in all directions.
Notation and description of the solution.
Let $S^{a,b}\subset S$ be the subset
$$
S^{a,b} = \{ x_1,\ldots,x_a, y_1, \ldots, y_b, z_1, \ldots, z_{d-2}\}
\subset S
$$
with $1\leq a,b\leq3$. It has $d+a+b-2$ elements. Denote by
$P_{(i,j)}^{a,b}$, with $i\leq a$ and $j\leq b$, the set of permutations of
$S^{a,b}$ with exactly $i$ adjacent $x$'s and $j$ adjacent $y$'s,
called permutations of type $(i,j)$.
Clearly the set of all permutations of $S^{a,b}$ is
$$
\bigcup_{\substack{1\leq i\leq a\\1\leq j\leq b}}
P_{(i,j)}^{a,b}.
$$
We want to compute $N_{(1,1)}^{3,3}$, the cardinal of
$P_{(1,1)}^{3,3}$. The idea is to do the computation recursively using
the following filtration of $S^{3,3}$:
$$
S^{1,1} \subset S^{2,1} \subset S^{3,1}
\subset S^{3,2} \subset S^{3,3}.
$$
Each $N_{(i,j)}^{a,b}$ corresponding to the subset $S^{a,b}$ in the
filtration is expressed as a linear combination with integer
coefficients of all the $N$'s of the previous subset. Once the
coefficients are determined (using the combinatorics of the
construction) for all these passages, the problem is solved.
Finding the coefficients.
Let's start with what we are looking for:
$$
N_{(1,1)}^{3,3} = d\,N_{(1,1)}^{3,2} + N_{(2,1)}^{3,2}.
$$
The other coefficients are all zero in this case. The explanation is
that a permutation of $S^{3,3}$ of type $(1,1)$ can only be issued
from a permutation of $S^{3,2}$ which is either of type $(1,1)$ or of
type $(2,1)$. For a permutation in the former case, there are
exactly $d=d+4-4$ positions (out of $d+4$) where $y_3$ can be inserted
(adjacent to neither $y_1$ nor $y_2$). In the latter case, $y_3$ must
be inserted in-between the two adjacent $x$'s.
We pursue with the formulas for $N_{(1,1)}^{3,2}$ and $N_{(2,1)}^{3,2}$
using the passage from $S^{3,1}$ to $S^{3,2}$ in our filtration. The
formula for $N_{(1,1)}^{3,2}$ is almost the same as the previous one:
$$
N_{(1,1)}^{3,2} = (d+3-2)\,N_{(1,1)}^{3,1} + N_{(2,1)}^{3,1}
= (d+1)\,N_{(1,1)}^{3,1} + N_{(2,1)}^{3,1}.
$$
The formula for $N_{(2,1)}^{3,2}$ is trickier. It reads
$$
N_{(2,1)}^{3,2}
= 0\,N_{(1,1)}^{3,1} + (d+3-3)\,N_{(2,1)}^{3,1} + 2N_{(3,1)}^{3,1} \\
= d\,N_{(2,1)}^{3,1} + 2N_{(3,1)}^{3,1}.
$$
The coefficient $d+3-3$ comes form the following observation: If a
type $(2,1)$ permutation of $S^{3,2}$ is issued form a type $(2,1)$
permutation of $S^{3,1}$, then $y_2$ has exactly $d+3-3$ possible
places to be inserted in: $d+3$ is the total number of places and $y_2$ can take
neither the place in-between the adjacent $x$'s nor any of the two places adjacent to
$y_1$.
Looking at the last two formulas, we note that from now on we need all
the $N$'s corresponding to the remaining subsets of the filtration.
But they are fewer and the formulas easier to figure out. We obtain
successively:
for $S^{2,1}\subset S^{3,1}$
$$
N_{(1,1)}^{3,1} = (d+2-4)\,N_{(1,1)}^{2,1} + 0\,N_{(2,1)}^{2,1}
$$
$$
N_{(2,1)}^{3,1} = 4\,N_{(1,1)}^{2,1} + (d+2-3)\,N_{(2,1)}^{2,1}
$$
$$
N_{(3,1)}^{3,1} = 0\,N_{(1,1)}^{2,1} + 3\,N_{(2,1)}^{2,1}
$$
for $S^{1,1}\subset S^{2,1}$
$$
N_{(1,1)}^{2,1} = (d+1-2)\,N_{(1,1)}^{1,1}
\quad\text{and}\quad
N_{(2,1)}^{2,1} = 2\,N_{(1,1)}^{1,1}.
$$
The computations
Now, to obtain $N_{(1,1)}^{3,3}$ it is sufficient to go backward
starting with $N_{(1,1)}^{1,1}=d!$. Let's perform the computations by
stressing their algorithmic character. I hope the notation below is
self-explanatory.
For $S^{2,1}$:
$$
\begin{array}{c||c|cr}
& (1,1) && result/d! \\ \hline
(1,1) & d-1 && (d-1) \\
(2,1) & 2 && 2
\end{array}
$$
For $S^{3,1}$:
$$
\begin{array}{c||c|c|cr}
& (1,1) & (2,1) && result/d! \\ \hline
(1,1) & d-2 & 0 && (d-2)(d-1) \\
(2,1) & 4 & d-1 && 6(d-1) \\
(3,1) & 0 & 3 && 6
\end{array}
$$
For $S^{3,2}$ (partial table):
$$
\begin{array}{c||c|c|c|cr}
& (1,1) & (2,1) & (3,1) && result/d! \\ \hline
(1,1) & d+1 & 1 & 0 && (d-1)(d^2-d+4) \\
(2,1) & 0 & d & 2 && 6(d^2-d+2)
\end{array}
$$
So
$$
\begin{split}
N_{(1,1)}^{3,3}
&= d\,N_{(1,1)}^{3,2} + N_{(2,1)}^{3,2} \\
&= \big( (d^2-d)(d^2-d+4) + 6(d^2-d+2) \big)\,d! \\
&= \big( (d^2-d)^2 + 10(d^2-d) + 12 \big)\,d! \\
&= \big( (d^2-d+4)(d^2-d+6) - 12 \big)\,d!.
\end{split}
$$
In particular, the answer to the initial question is
$$
\frac{(5^2-5+4)(5^2-5+6)-12}{3!\,3!}\,\frac{5!}{3!}
= 20\,\frac{2\cdot26-1}{3} = 340.
$$
