How do I find the 8th term for the recursive sequence list in the body? $3,0,2,3,2,5$ and I know the $8$th term is $10$.

If we know that the recursive sequence is a simple linear recurrence, then an $n^{th}$ order linear recurrence needs $2n$ known values - the initial $n$ terms and the coefficients.
You have seven knowns, so we can hypothesize that the recurrence has the form
$$
a_n=\alpha_1a_{n-1}+\alpha_2a_{n-2}+\alpha_3a_{n-3}
$$
with initial terms
$$
\begin{align}
a_0 &= 3 \\
a_1 &= 0 \\
a_2 &= 2.
\end{align}
$$
With the next three terms we can create a set of linear equations:
$$
\begin{align}
a_3 &= 3 &= \alpha_1(2)+\alpha_2(0)+\alpha_3(3) \\
a_4 &= 2 &= \alpha_1(3)+\alpha_2(2)+\alpha_3(0) \\
a_5 &= 5 &= \alpha_1(2)+\alpha_2(3)+\alpha_3(2).
\end{align}
$$
Solving this yields
$$
\begin{align}
\alpha_1 &= 0 \\
\alpha_2 &= 1 \\
\alpha_3 &= 1.
\end{align}
$$
Now we can calculate all terms, specifically $a_6=5$, $a_7=7$, and $a_8=10$. So the Perrin Sequence as mentioned is the only third-order simple linear recurrence given your constraints. 
