Given that $\alpha \beta \gamma = 45$ and $\alpha, \beta, \gamma \in \mathbb{R}$, consider the equation $x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$ for some $x, y, z \in \mathbb{R}$ where $x y z \neq 0$. To find the value of $6 \alpha + 4 \beta + \gamma$, follow these steps:

1. Express the given equation in matrix form:

$ \begin{aligned} & \alpha x + y + 2 z = 0 \\ & x + \beta y + 3 z = 0 \\ & 2 x + 2 y + \gamma z = 0 \end{aligned} $

1. Since $x, y, z \neq 0$, the determinant of the coefficients matrix must be zero:

$ \left|\begin{array}{ccc} \alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma \end{array}\right| = 0 $

1. Calculate the determinant of the matrix:

$ \alpha \beta \gamma - 6 \alpha - 4 \beta - \gamma + 10 = 0 $

1. Given $\alpha \beta \gamma = 45$, substitute this value into the equation:

$ 45 - 6 \alpha - 4 \beta - \gamma + 10 = 0 $

1. Simplify the equation:

$ 45 + 10 = 6 \alpha + 4 \beta + \gamma $

1. Thus,

$ 6 \alpha + 4 \beta + \gamma = 55 $

So, the value of $6 \alpha + 4 \beta + \gamma$ is 55.