5. The positive sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
\frac{1}{a_{n} a_{n+1}}+\frac{1}{a_{n} a_{n+2}}+\frac{1}{a_{n+1} a_{n+2}}=1\left(n \in \mathbf{Z}_{+}\right), \\
a_{1}+a_{3}=6,
\end{array}
$$
$a_{1}, a_{2}, a_{3}$ are in increasing geometric progression. If $S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$, then the value of $\left[S_{2014}\right]$ is ( ) ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
(A) 5368
(B) 5367
(C) 5363
(D) 5362
5. A.

From the given, we have
$$
a_{n}+a_{n+1}+a_{n+2}=a_{n} a_{n+1} a_{n+2} \text {. }
$$

Thus, $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}$.
And $a_{1} a_{3}=a_{2}^{2}$, so
$$
\begin{array}{l}
a_{2}^{3}=a_{2}+6 \\
\Rightarrow\left(a_{2}-2\right)\left(a_{2}^{2}+2 a_{2}+3\right)=0 \\
\quad \Rightarrow a_{2}=2 .
\end{array}
$$

Since $a_{1}, a_{2}, a_{3}$ form a geometric sequence,
$$
\Rightarrow a_{1}+\frac{4}{a_{1}}=6 \Rightarrow a_{1}=3 \pm \sqrt{5} \text {. }
$$

Given that $a_{1}, a_{2}, a_{3}$ are monotonically increasing, we have $a_{1}=3-\sqrt{5}$.
Combining with equation (1), we get
$$
a_{n+1}+a_{n+2}+a_{n+3}=a_{n+1} a_{n+2} a_{n+3} \text {. }
$$
Subtracting (1) from (2) yields
$$
\left(a_{n+3}-a_{n}\right)\left(1-a_{n+1} a_{n+2}\right)=0 \text {. }
$$

Thus, $a_{n}=a_{n+3}$, meaning $\left\{a_{n}\right\}$ is a periodic sequence with a period of 3.
$$
\begin{array}{l}
\text { Then } a_{2014}=a_{1}=3-\sqrt{5} . \\
\text { Hence } S_{2014}=S_{2013}+a_{2014} \\
=\frac{2013}{3}(6+2)+3-\sqrt{5} \\
=5368+(3-\sqrt{5}) .
\end{array}
$$

Therefore, $\left[S_{2014}\right]=5368$.