Given \( \alpha \beta \gamma = 45 \), \( \alpha, \beta, \gamma \in \mathbb{R} \),
\[ x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0). \]
Expanding, we get:
\[ x \alpha + y + 2z = 0, \] \[ x + y \beta + 3z = 0, \] \[ 2x + 2y + z \gamma = 0. \]
Since \( xyz \neq 0 \), the determinant of the coefficient matrix must be zero for a non-trivial solution:
\[ \begin{vmatrix} \alpha & 1 & 2 \\ 1 & \beta & 2 \\ 2 & 3 & \gamma \end{vmatrix} = 0. \]
Expanding the determinant:
\[ \alpha \begin{vmatrix} \beta & 2 \\ 3 & \gamma \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 2 & \gamma \end{vmatrix} + 2 \begin{vmatrix} 1 & \beta \\ 2 & 3 \end{vmatrix} = 0. \]
Calculating each minor:
\[ \begin{vmatrix} \beta & 2 \\ 3 & \gamma \end{vmatrix} = \beta \gamma - 6, \] \[ \begin{vmatrix} 1 & 2 \\ 2 & \gamma \end{vmatrix} = \gamma - 6, \] \[ \begin{vmatrix} 1 & \beta \\ 2 & 3 \end{vmatrix} = 3 - 2\beta. \]
Substituting:
\[ \alpha (\beta \gamma - 6) - (\gamma - 6) + 2(3 - 2\beta) = 0. \]
Simplify:
\[ \alpha \beta \gamma - 6\alpha - \gamma + 6 + 6 - 4\beta = 0. \]
Since \( \alpha \beta \gamma = 45 \):
\[ 45 - 6\alpha - \gamma + 12 - 4\beta = 0. \]
Rearranging:
\[ 6\alpha + 4\beta + \gamma = 55. \]
Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which
\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]
is equal to __________.