Solution:
Let $x_{n}$ be the number of ways in which $A$ can get back the ball after $n$ passes. Let $y_{n}$ be the number of ways in which the ball goes back to a fixed person other than $A$ after $n$ passes. Then
$$x_{n}=3 y_{n-1},$$
and
$$y_{n}=x_{n-1}+2 y_{n-1}$$
We also have $x_{1}=0$, $x_{2}=3$, $y_{1}=1$ and $y_{2}=2$.

Eliminating $y_{n}$ and $y_{n-1}$, we get $x_{n+1}=3 x_{n-1}+2 x_{n}$. Thus
$$\begin{aligned} & x_{3}=3 x_{1}+2 x_{2}=2 \times 3=6 \\ & x_{4}=3 x_{2}+2 x_{3}=(3 \times 3)+(2 \times 6)=9+12=21 \\ & x_{5}=3 x_{3}+2 x_{4}=(3 \times 6)+(2 \times 21)=18+42=60 \\ & x_{6}=3 x_{4}+2 x_{5}=(3 \times 21)+(2 \times 60)=63+120=183 \\ & x_{7}=3 x_{5}+2 x_{6}=(3 \times 60)+(2 \times 183)=180+366=546 \end{aligned}$$

Alternate solution:
Since the ball goes back to one of the other 3 persons, we have
$$x_{n}+3 y_{n}=3^{n}$$
since there are $3^{n}$ ways of passing the ball in $n$ passes. Using $x_{n}=3 y_{n-1}$, we obtain
$$x_{n-1}+x_{n}=3^{n-1}$$
with $x_{1}=0$. Thus
$$\begin{array}{r} x_{7}=3^{6}-x_{6}=3^{6}-3^{5}+x_{5}=3^{6}-3^{5}+3^{4}-x_{4}=3^{6}-3^{5}+3^{4}-3^{3}+x_{3} \\ =3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-x_{2}=3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-3 \\ =\left(2 \times 3^{5}\right)+\left(2 \times 3^{3}\right)+(2 \times 3)=486+54+6=546 \end{array}$$