There are $\binom{6}{3} = 20$ sequences of three $A$'s and three $B$'s. Consider the ten sequences of three $A$'s and three $B$'s that begin with an $A$.
$\color{red}{AAABBB}$
$\color{green}{AABABB}$
$\color{green}{AABBAB}$
$\color{blue}{AABBBA}$
$\color{green}{ABAABB}$
$\color{cyan}{ABABAB}$
$\color{magenta}{ABABBA}$
$\color{green}{ABBAAB}$
$\color{magenta}{ABBABA}$
$\color{blue}{ABBBAA}$
No matter how we place $C$'s in the sequence $\color{red}{AAABBB}$, at least two consecutive letters will be the same.
There is only one way to place the $C$'s in the sequences $\color{blue}{AABBBA}$ and $\color{blue}{ABBBAA}$ since we are forced to place a $C$ between each pair of consecutive $B$'s and the pair of consecutive $A$'s.
The number of ways we can fill three of the seven spaces (the beginning, the end, and the five spaces between consecutive letters) in the sequence $\color{cyan}{ABABAB}$ is $\binom{7}{3}$.
We must place a $C$ between the pair of consecutive $B$'s in the sequences $\color{magenta}{ABABBA}$ and $\color{magenta}{ABBABA}$, which leaves us $\binom{6}{2}$ ways to insert the two remaining $C$'s in the six remaining spaces.
We must place one $C$ between the pair of consecutive $A$'s and another $C$ between the pair of consecutive $B$'s in the sequences $\color{green}{AABABB}$, $\color{green}{AABBAB}$, $\color{green}{ABAABB}$, $\color{green}{ABBAAB}$, leaving five spaces in which to place the remaining $C$.
Hence, the number of sequences of three $A$'s, three $B$'s, and three $C$'s that begin with $A$ that do not contain consecutive letters that are the same is
$$1 \cdot 0 + 2 \cdot 1 + 2 \cdot \binom{6}{2} + 4 \cdot \binom{5}{1} + 1 \cdot \binom{7}{3} = 0 + 2 + 30 + 20 + 35 = 87$$
By symmetry, there are also $87$ such sequences that begin with a $B$. Hence, the number of sequences of three $A$'s, three $B$'s, and three $C$'s that do not contain consecutive letters that are the same is $2 \cdot 87 = 174$.
Your method both under and over counts. It undercounts by neglecting patterns like $GGBBGB$, it overcounts by both permuting and then repopulating patterns. If, instead, you skip the permutations (which can all be realized through the later repopulation) you would get $2^3\times 3!\times 3!=288$ which is the correct answer (if you disallow the two patterns $GGBBGB,\;BGBBGG$)
The hard part is to get all the possible patterns. I'll do that here:
By looking at the possible strings,starting with $B=boy$, there are five possible types of arrangements: $$\{BGGBGB,\;BGGBBG,\;BGBGBG,\;BGBGGB,\;BGBBGG\}$$
Starting with $G=girl$ we also have five possible arrangements, namely $$\{GBBGBG,\;GBBGGB,\;GBGBGB,\;GBGBBG,\;GGBBGB\}$$
Each of these can be populated in $3!\times 3!=36$ ways, hence $360$.
To elaborate on the counting:
Case I. say we start with $B$. Then the next has to be $G$.
Ia. $BGG...$. Fourth must be $B$, whence $BGGB$ but then either $\fbox {BGGBBG}$ or $\fbox {BGGBGB}$ work.
Ib. $BGB...$. We could have $BGBB$ in which case we must have $\fbox {BGBBGG}$ or we could have $BGBG$ win which case we could have either $\fbox {BGBGGB}$ or $\fbox{BGBGBG}$.
Case II (starting with $G$) is similar.
Best Answer
See we will go by your method using complements. See total ways are $720$ . now lets group three men so let us assume them $x$ . now let us place them in positions $123$ so total ways are $3!.3!=36$ as men and women can e arranged amongst in $3!$ ways. Now place $x$ at place $234$ again same number of ways ie $36$ now similarly at $345,456$ so total ways become $36\cdot 4=144$ now group any two men . this can be done in $3$ ways. Place them at place $1,2$ now total ways become $3.2!.3.3!=108$ . men can be arranged in $2!$ ways $3rd$ place has to be female or it becomez similar to $3$ men together so $3$ ways and $1M,2w$ can be arranged in $3!$ ways. Note that now they cant be placed in positions $(23),(34),(45)$ as it becomes similar to $3$ men together. So now they can be placed at $56$ position. Again similar to $12$ position we get $108$ ways . now remember we are talking complement of total permitted ways . hence total ways are $720-144-216=360$