Let $$\alpha\beta\gamma = 45$$; $$\alpha, \beta, \gamma \in \mathbb{R}$$. If $$x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$$ for some $$x, y, z \in \mathbb{R}, xyz \neq 0$$, then $$6\alpha + 4\beta + \gamma$$ is equal to _______

$$x(\alpha,1,2)+y(1,\beta,2)+z(2,3,\gamma)=(0,0,0)$$ has nontrivial solution, so determinant = 0:

$$\begin{vmatrix}\alpha&1&2\\1&\beta&2\\2&3&\gamma\end{vmatrix} = 0$$.

$$\alpha(\beta\gamma-6)-1(\gamma-4)+2(3-2\beta) = 0$$.

$$\alpha\beta\gamma - 6\alpha - \gamma + 4 + 6 - 4\beta = 0$$.

Since $$\alpha\beta\gamma = 45$$: $$45 - 6\alpha - \gamma - 4\beta + 10 = 0 \Rightarrow 6\alpha + 4\beta + \gamma = 55$$.

The answer is 55.