This problem requires a strong grasp of the properties of determinants and adjoints of matrices. We will systematically break down the given determinant expressions using these properties to solve for the unknown values.

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Key Concepts and Formulas Used (for a square matrix $X$ of order $n=3$ and a scalar $k$):

1. Determinant of a scalar multiple: $\det(kX) = k^n \det(X) = k^3 \det(X)$
2. Determinant of an adjoint matrix: $\det(\operatorname{adj}(X)) = (\det X)^{n-1} = (\det X)^2$
3. Adjoint of a scalar multiple: $\operatorname{adj}(kX) = k^{n-1} \operatorname{adj}(X) = k^2 \operatorname{adj}(X)$
4. Adjoint of an adjoint matrix: $\operatorname{adj}(\operatorname{adj}(X)) = (\det X)^{n-2} X = (\det X) X$ (since $n-2 = 1$)

Let $D = \det A$ for simplicity. Since $A$ is a non-singular matrix, $D \neq 0$.

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Step-by-Step Solution:

We are given two equations involving determinants and need to find the value of $|3m+2n|$.

Part 1: Analyzing the First Equation to find $\det A$

The first equation is $\det(3 \operatorname{adj}(2 \operatorname{adj}((\det A) A)))=3^{-13} \cdot 2^{-10}$.

Let's simplify the expression from the innermost part outwards:

1. Innermost term: $(\det A) A = DA$. This is a scalar $D$ multiplied by matrix $A$.
2. Second innermost determinant (argument of outer adj): $\operatorname{adj}(DA)$ Using property 3, $\operatorname{adj}(kX) = k^2 \operatorname{adj}(X)$, with $k=D$ and $X=A$: $\operatorname{adj}(DA) = D^2 \operatorname{adj}(A)$
3. Third innermost determinant (argument of outermost adj): $2 \operatorname{adj}(DA)$ Substitute the result from step 2: $2(D^2 \operatorname{adj}(A)) = 2D^2 \operatorname{adj}(A)$
4. Outermost adjoint: $\operatorname{adj}(2D^2 \operatorname{adj}(A))$ Using property 3 again, $\operatorname{adj}(kX) = k^2 \operatorname{adj}(X)$, with $k=2D^2$ and $X=\operatorname{adj}(A)$: $\operatorname{adj}(2D^2 \operatorname{adj}(A)) = (2D^2)^2 \operatorname{adj}(\operatorname{adj}(A))$ $= 4D^4 \operatorname{adj}(\operatorname{adj}(A))$ Now, using property 4, $\operatorname{adj}(\operatorname{adj}(A)) = DA$: $= 4D^4 (DA) = 4D^5 A$
5. Argument of the outermost determinant: $3 \operatorname{adj}(2 \operatorname{adj}(DA))$ Substitute the result from step 4: $3(4D^5 A) = 12D^5 A$
6. Final determinant evaluation: $\det(12D^5 A)$ Using property 1, $\det(kX) = k^3 \det(X)$, with $k=12D^5$ and $X=A$: $\det(12D^5 A) = (12D^5)^3 \det(A)$ $= 12^3 (D^5)^3 D = 12^3 D^{15} D = 12^3 D^{16}$ We know $12 = 2^2 \cdot 3$, so $12^3 = (2^2 \cdot 3)^3 = 2^6 \cdot 3^3$. Thus, the LHS simplifies to $2^6 \cdot 3^3 \cdot D^{16}$.

Now, equate this to the given value: $2^6 \cdot 3^3 \cdot D^{16} = 3^{-13} \cdot 2^{-10}$ To find $D^{16}$, divide both sides by $2^6 \cdot 3^3$: $D^{16} = \frac{3^{-13} \cdot 2^{-10}}{2^6 \cdot 3^3}$ $D^{16} = 2^{-10-6} \cdot 3^{-13-3}$ $D^{16} = 2^{-16} \cdot 3^{-16}$ $D^{16} = (2 \cdot 3)^{-16} = 6^{-16}$ From this, $D = \det A = \pm 6^{-1}$. For the subsequent calculations, $D^2$ will be used, which is unique: $D^2 = (6^{-1})^2 = 6^{-2}$

Part 2: Analyzing the Second Equation to find $m$ and $n$

The second equation is $\det(3\operatorname{adj}(2 \mathrm{A}))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}$.

Let's simplify the expression from the innermost part outwards:

1. Innermost term (argument of adj): $2A$
2. Adjoint term: $\operatorname{adj}(2A)$ Using property 3, $\operatorname{adj}(kX) = k^2 \operatorname{adj}(X)$, with $k=2$ and $X=A$: $\operatorname{adj}(2A) = 2^2 \operatorname{adj}(A) = 4 \operatorname{adj}(A)$
3. Argument of the outermost determinant: $3 \operatorname{adj}(2A)$ Substitute the result from step 2: $3(4 \operatorname{adj}(A)) = 12 \operatorname{adj}(A)$
4. Final determinant evaluation: $\det(12 \operatorname{adj}(A))$ Using property 1, $\det(kX) = k^3 \det(X)$, with $k=12$ and $X=\operatorname{adj}(A)$: $\det(12 \operatorname{adj}(A)) = 12^3 \det(\operatorname{adj}(A))$ Using property 2, $\det(\operatorname{adj}(A)) = (\det A)^2 = D^2$: $= 12^3 D^2$ Substitute $12^3 = 2^6 \cdot 3^3$: $= 2^6 \cdot 3^3 \cdot D^2$ Now, substitute the value of $D^2 = 6^{-2}$ found in Part 1: $= 2^6 \cdot 3^3 \cdot (6^{-2})$ Since $6^{-2} = (2 \cdot 3)^{-2} = 2^{-2} \cdot 3^{-2}$: $= 2^6 \cdot 3^3 \cdot 2^{-2} \cdot 3^{-2}$ Combine the powers of 2 and 3: $= 2^{6-2} \cdot 3^{3-2}$ $= 2^4 \cdot 3^1$

This expression is given as $2^m \cdot 3^n$. Therefore, by comparing the powers: $m=4$ $n=1$

Part 3: Calculating the Final Value

We need to find $|3m+2n|$. Substitute the values of $m$ and $n$: $|3m+2n| = |3(4) + 2(1)|$ $= |12 + 2|$ $= |14|$ $= 14$

The final answer is $\boxed{14}$.

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Summary and Key Takeaway:

This problem demonstrates the importance of meticulously applying the properties of determinants and adjoints. By systematically simplifying the expressions from the innermost operations outwards, we avoided errors. The key properties for an $n \times n$ matrix are $\det(kX) = k^n \det(X)$ and $\det(\operatorname{adj}(X)) = (\det X)^{n-1}$. Additionally, the property $\operatorname{adj}(kX) = k^{n-1} \operatorname{adj}(X)$ is crucial for simplifying complex adjoint arguments. Careful tracking of powers of scalars is essential to arrive at the correct final answer.