Let A be a matrix of order $$3 \times 3$$ and $$|A| = 5$$. If $$|2\text{adj}(3A \text{adj}(2A))| = 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}$$, $$\alpha, \beta, \gamma \in \mathbb{N}$$ then $$\alpha + \beta + \gamma$$ is equal to

We are given a $$3 \times 3$$ matrix $$A$$ with $$|A| = 5$$ and we must evaluate
$$|\,2\,\operatorname{adj}\!\bigl(3A \,\operatorname{adj}(2A)\bigr)|$$.

Step 1: Pull out the scalar “2”.
For a $$3 \times 3$$ matrix $$M$$, $$|kM| = k^{3}\,|M|$$. Hence
$$|\,2\,\operatorname{adj}(3A \operatorname{adj}(2A))| = 2^{3}\,|\operatorname{adj}(3A \operatorname{adj}(2A))|.$$

Step 2: Determinant of an adjugate.
For any $$3 \times 3$$ matrix $$C$$, $$|\operatorname{adj}C| = |C|^{3-1} = |C|^{2}$$.
Thus
$$|\operatorname{adj}(3A \operatorname{adj}(2A))| = \bigl|\,3A \operatorname{adj}(2A)\bigr|^{2}.$$

Step 3: Expand the determinant inside.
Using $$|PQ| = |P|\,|Q|$$,
$$\bigl|\,3A \operatorname{adj}(2A)\bigr| = |\,3A| \; \cdot \; |\operatorname{adj}(2A)|.$$

Step 4: Evaluate $$|\,3A|$$.
Again, $$|kA| = k^{3}|A| \implies |\,3A| = 3^{3}\,|A| = 27 \times 5 = 3^{3}\cdot 5.$$

Step 5: Evaluate $$|\operatorname{adj}(2A)|$$.
First find $$|\,2A| = 2^{3}|A| = 8 \times 5 = 2^{3}\cdot 5.$}
Then $$|\operatorname{adj}(2A)| = |\,2A|^{2} = (2^{3}$$\cdot$$ 5)^{2} = 2^{6}$$\cdot$$ 5^{2}.$$

Step 6: Combine results of Steps 4 and 5.
$$\bigl|\,3A \operatorname{adj}(2A)\bigr| = (3^{3}$$\cdot$$ 5)\,(2^{6}$$\cdot$$ 5^{2}) = 2^{6}\,3^{3}\,5^{3}.$$

Step 7: Square the determinant (per Step 2).
$$\bigl|\,3A \operatorname{adj}(2A)\bigr|^{2} = \bigl(2^{6}\,3^{3}\,5^{3}\bigr)^{2} = 2^{12}\,3^{6}\,5^{6}.$$

Step 8: Multiply by the outer factor $$2^{3}$$ (from Step 1).
Final value:
$$|\,2\,\operatorname{adj}(3A \operatorname{adj}(2A))| = 2^{3}\,(2^{12}\,3^{6}\,5^{6}) = 2^{15}\,3^{6}\,5^{6}.$$

Step 9: Identify the exponents.
$$$$\alpha$$ = 15,\; $$\beta$$ = 6,\; $$\gamma$$ = 6 \;\;\Longrightarrow\;\; $$\alpha + \beta + \gamma$$ = 15 + 6 + 6 = 27.$$

Hence the required sum is $$27$$, which corresponds to Option C.