first column of $A^{-1}$ is

$$ \begin{bmatrix} \gamma \\ \alpha \\ \beta \end{bmatrix} $$

So when we multiply $A$ by this column, we should get the first column of the identity matrix:

$$ A \begin{bmatrix} \gamma \\ \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$

Now,

$$ \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} \gamma \\ \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} 2\gamma - \beta \\ 5\gamma + \alpha \\ \alpha + 3\beta \end{bmatrix} $$

So we get

$$ 2\gamma - \beta = 1 $$

$$ 5\gamma + \alpha = 0 $$

$$ \alpha + 3\beta = 0 $$

From

$$ 5\gamma + \alpha = 0 $$

we get

$$ \alpha = -5\gamma $$

From

$$ \alpha + 3\beta = 0 $$

we get

$$ \alpha = -3\beta $$

So,

$$ -5\gamma = -3\beta $$

$$ \beta = \frac{5\gamma}{3} $$

Now use

$$ 2\gamma - \beta = 1 $$

Substitute $\beta = \frac{5\gamma}{3}$:

$$ 2\gamma - \frac{5\gamma}{3} = 1 $$

$$ \frac{6\gamma - 5\gamma}{3} = 1 $$

$$ \frac{\gamma}{3} = 1 $$

$$ \gamma = 3 $$

Then,

$$ \alpha = -5\gamma = -15 $$

and

$$ \beta = \frac{5\gamma}{3} = 5 $$

Now,

$$ |\alpha \cdot \beta \cdot \gamma| = |(-15)(5)(3)| $$

$$ = |-225| $$

$$ = 225 $$

So the correct option is

$$ \boxed{225} $$

Option C.

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