1. Use the determinant scaling property

For a $3 \times 3$ matrix $A$, $|2A|=2^3|A|=8|A|.$

Given $|2A|^3=2^{21},$ so $(8|A|)^3=2^{21}.$

Now, $8^3|A|^3=2^{21}$ $2^9|A|^3=2^{21}$ $|A|^3=2^{12}.$

Hence, $|A|=2^4=16$ because determinant is an integer here.

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2. Find the determinant of $A$

The matrix is $A=\begin{bmatrix}1&0&0\\0&\alpha&\beta\\0&\beta&\alpha\end{bmatrix}.$

Since the first row/column isolates a $1$, we get

\alpha^2-\beta^2.$$ So, $$\alpha^2-\beta^2=16.$$ This gives $$(\alpha-\beta)(\alpha+\beta)=16.$$ Since $\alpha,\beta\in \mathbb{Z}$, both factors are integers. --- 3. **Check the options** We need a value of $\alpha$ for which $$\alpha^2-\beta^2=16$$ for some integer $\beta$. Equivalently, $$\beta^2=\alpha^2-16.$$ Now test options: - **A: $\alpha=9$** $$\beta^2=81-16=65,$$ not a perfect square. - **B: $\alpha=17$** $$\beta^2=289-16=273,$$ not a perfect square. - **C: $\alpha=3$** $$\beta^2=9-16=-7,$$ impossible. - **D: $\alpha=5$** $$\beta^2=25-16=9,$$ so $\beta=\pm 3$, which is an integer. Thus $\alpha=5$ is possible. --- 4. **Final answer** The correct option is $$\boxed{5}.$$