The answer is the coefficient of the $x^6$ term in $$\begin{align*}f(x)=\ &(x^{-1} + 1 + x + x^2)(x + x^2 + x^3+x^4)^2\\=\ &x\left(1+ x+x^2 + ^3\right)^3\end{align*}$$

which is the same as the coefficient of the $x^5$ term in $$\begin{align*}g(x)=\ &\left(1+x+x^2+x^3\right)^3\\=\ &\frac{1}{(1-x)^3}\cdot(1-x^4)^3\\=\ &\left(\sum_{k=0}^{\infty}\binom{-3}{k}(-1)^kx^k\right)\left(1-3x^4+3x^8-x^{12}\right)\end{align*}$$

Therefore, the answer is $$\binom{-3}{5}(-1)^5-3\cdot\binom{-3}{1}(-1)^1=\boxed{12}$$

This result can be verified by listing all the qualified solutions:

$$\begin{align*} &(-1, 3, 4),\ (-1, 4, 3),\ (0, 2, 4),\ (0,3,3),\ (0,4, 2),(1, 1, 4),\\ &(1, 2, 3),\ (1, 3, 2),\ (1, 4, 1),\ (2, 1, 3),\ (2, 2, 2),\ (2, 3, 1) \end{align*}$$