Study the sequence $a_n - a_{n-1}$. It goes like $17, 102, 612$ etc. Do you see that this is a geometric series, with first term $17$ and constant ratio $6$?
Hence, $a_n- a_{n-1} = 17 \cdot 6^n$, you could say? Now figure out the general formula for $a_n$, and see if the additions work out. I'l give the answer in this "hidden hint":
$a_n = \dfrac{17\cdot 6^n - 2}{5}$. The sum, up to $n$ terms, is $\dfrac{17\cdot 6^{n+1} -10n-27}{25}$.
The following is a step-by-step solution which reduces the problem to a linear homogeneous recurrence with constant coefficients, which can be solved with standard techniques.
The exponential term can be "absorbed" into the recurrence by dividing by $\,3^{n-2}\,$ and defining $u_n = \dfrac{a_n}{3^n}\,$:
$$
\begin{align}
a_n=a_{n-1}+6a_{n-2}+3^n \quad&\iff\quad 9 \cdot \frac{a_n}{3^n} = 3 \cdot \frac{a_{n-1}}{3^{n-1}} + 6 \cdot \frac{a_{n-2}}{3^{n-2}} + 9
\\ &\iff\quad 9 u_n = 3u_{n-1}+6u_{n-2}+ 9
\\ &\iff\quad 3 u_n = u_{n-1} + 2u_{n-2} + 3 \tag{1}
\end{align}
$$
The constant term can then be "absorbed" by trying a constant translation $\,u_n = v_n + \lambda\,$ in $(1)\,$:
$$
\require{cancel}
\begin{align}
3(v_n + \cancel{\lambda}) = (v_{n-1}+\cancel{\lambda}) + 2(v_{n-2}+\cancel{\lambda}) + 3 \quad\iff\quad 3 v_n = v_{n-1} + 2v_{n-2} + 3 \tag{2}
\end{align}
$$
The resulting recurrence $(2)$ is identical to $(1)$, so this substitution does not work because the constant term cannot be eliminated this way (which actually happens because the characteristic polynomial has $1$ as a root). Next try in such cases is a linear transformation $u_n = v_n + \lambda n\,$:
$$
\require{cancel}
\begin{align}
3(v_n+\bcancel{\lambda n}) &= \big(v_{n-1}+ \lambda(\bcancel{n}-1)\big) + 2\big(v_{n-2}+\lambda(\bcancel{n}-2)\big) + 3
\\ \iff\quad\quad\quad\quad 3 v_n &= v_{n-1} + 2 v_{n-2} - 5\lambda + 3 \tag{3}
\end{align}
$$
This one works, by choosing $\lambda = \dfrac{3}{5}$ to cancel the constant term in $(3)$, so $\,u_n=v_n + \dfrac{3}{5}n\,$ satisfies:
$$
3 v_n = v_{n-1} + 2 v_{n-2} \tag{4}
$$
The latter is a linear homogeneous recurrence with constant coefficients. The characteristic polynomial is $3t^2-t-2$ with roots $\big\{1, -\frac{2}{3}\big\}$, so $v_n = c_1 + c_2 \cdot \dfrac{(-1)^n2^n}{3^n}$ for some constants $c_1, c_2$. By reversing the substitutions, it follows that:
$$
a_n = 3^n u_n = 3^n\left(v_n+\frac{3}{5}n\right) = c_1 \cdot 3^n + c_2 \cdot (-1)^n2^n + \frac{1}{5} \cdot n \cdot 3^{n+1} \tag{5}
$$
Finally, the constants $c_1,c_2$ can be determined from the initial conditions by solving the system:
$$
\begin{cases}
\begin{align}
5 = a_0 &= c_1 + c_2
\\ 0 = a_1 &= 3c_1 - 2c_2 + \frac{9}{5}
\end{align}
\end{cases}
$$
Best Answer
I'm not sure what process you used, as different processes generalize differently to this case. I'm personally partial to generating functions, so I'll discuss that method.
The key observation is that coefficients $(n-1)a_{n-2}$ look a lot like a derivative. After all, if we let
$$A = \sum a_n x^n$$
Then $$\frac{d}{dx} A = \sum n a_n x^{n-1}$$.
In the example you posed, $a_n = (n-1)a_{n-2} + na_{n-3}$, we could multiply everything in sight by $x^n$ and sum to get
$$A = \sum a_n x^n = \sum \left ( (n-1) a_{n-2} x^n + n a_{n-3} x^n \right )$$
We split up the right hand side as
$$x^2 \sum (n-1) a_{n-2} x^{n-2} + x^3 \sum n a_{n-3} x^{n-3}$$
Now by using derivative rules, you can simplify this to get a differential equation for $A$. Then by solving the differential equation, you can get a closed form for $A$. All of this can be automated in a computer algebra system like sage, though you can do it by hand with some persistence.
You can read more about this technique in Wilf's fantastic book generatingfunctionology.
I hope this helps ^_^