This problem tests your understanding of the fundamental properties of determinants and adjoint matrices, which are crucial concepts in linear algebra for JEE Mathematics.

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Key Concepts and Formulas Used

Before we dive into the solution, let's recall the essential properties that will be applied:

For a square matrix $X$ of order $n$:

1. Determinant of Adjoint: The determinant of the adjoint of $X$ is given by: $|\text{adj}X| = |X|^{n-1}$ This formula is vital for simplifying expressions involving adjoints.

2. Determinant of Adjoint of Adjoint: Extending the above property, the determinant of the adjoint of the adjoint of $X$ is: $|\text{adj}(\text{adj}X)| = |X|^{(n-1)^2}$ This is a direct application of the first property, where $X$ is replaced by $\text{adj}X$. Since $\text{adj}X$ is also an $n \times n$ matrix, $|\text{adj}(\text{adj}X)| = |\text{adj}X|^{n-1} = (|X|^{n-1})^{n-1} = |X|^{(n-1)^2}$.

3. Determinant of Scalar Multiple of a Matrix: If $k$ is a scalar, then the determinant of $kX$ is: $|kX| = k^n |X|$ This property highlights how scalar multiplication affects the determinant, raising the scalar to the power of the matrix's order.

In this problem, the given matrix $A$ is a $3 \times 3$ matrix, so the order $n=3$.

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Step-by-Step Solution

1. Calculate the Determinant of Matrix A ($|A|$).

First, we need to find the determinant of the given matrix $A$. $A = \left[ {\begin{matrix} 1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & { - 1} & 3 \end{matrix} } \right]$

We will expand the determinant along the first row ($R_1$) for simplicity. $|A| = 1 \cdot \left| {\begin{matrix} 3 & 4 \\ { - 1} & 3 \end{matrix} } \right| - 1 \cdot \left| {\begin{matrix} 1 & 4 \\ 1 & 3 \end{matrix} } \right| + 2 \cdot \left| {\begin{matrix} 1 & 3 \\ 1 & { - 1} \end{matrix} } \right|$

Now, calculate the $2 \times 2$ determinants: $|A| = 1 \cdot ((3)(3) - (4)(-1)) - 1 \cdot ((1)(3) - (4)(1)) + 2 \cdot ((1)(-1) - (3)(1))$ $|A| = 1 \cdot (9 - (-4)) - 1 \cdot (3 - 4) + 2 \cdot (-1 - 3)$ $|A| = 1 \cdot (9 + 4) - 1 \cdot (-1) + 2 \cdot (-4)$ $|A| = 1 \cdot (13) + 1 - 8$ $|A| = 13 + 1 - 8$ $|A| = 14 - 8$ $|A| = 6$ So, the determinant of matrix A is $6$.

Why this step is taken: The problem asks for an expression involving $|A|$, so calculating $|A|$ is the foundational first step.

2. Express the Given Expression in Terms of A.

We are asked to find the value of $\frac{|\text{adj}B|}{|C|}$. We are given $B = \text{adj}A$ and $C = 3A$. Substitute these definitions into the expression: $\frac{|\text{adj}B|}{|C|} = \frac{|\text{adj}(\text{adj}A)|}{|3A|}$

Why this step is taken: By substituting $B$ and $C$ with their definitions in terms of $A$, we transform the problem into one that can be solved using the properties of determinant and adjoint related to matrix $A$, which is the known matrix.

3. Apply Determinant Properties to the Numerator.

The numerator is $|\text{adj}(\text{adj}A)|$. Using the property $|\text{adj}(\text{adj}X)| = |X|^{(n-1)^2}$ with $X=A$ and $n=3$: $|\text{adj}(\text{adj}A)| = |A|^{(3-1)^2}$ $|\text{adj}(\text{adj}A)| = |A|^{(2)^2}$ $|\text{adj}(\text{adj}A)| = |A|^4$

Why this step is taken: This step simplifies the complex expression involving nested adjoints into a straightforward power of $|A|$, making it much easier to work with.

4. Apply Determinant Properties to the Denominator.

The denominator is $|C| = |3A|$. Using the property $|kX| = k^n |X|$ with $X=A$, $k=3$, and $n=3$: $|3A| = 3^3 |A|$ $|3A| = 27 |A|$

Why this step is taken: This step simplifies the determinant of a scalar multiple of $A$ into a product of a constant and $|A|$, preparing it for combination with the numerator.

5. Combine and Simplify the Expression.

Now, substitute the simplified numerator and denominator back into the main expression: $\frac{|\text{adj}(\text{adj}A)|}{|3A|} = \frac{|A|^4}{3^3 |A|}$

We can simplify the powers of $|A|$: $\frac{|A|^4}{3^3 |A|} = \frac{|A|^{4-1}}{3^3}$ $= \frac{|A|^3}{3^3}$

Now, substitute the value of $|A|=6$ that we calculated in Step 1: $= \frac{6^3}{3^3}$

This can be written as: $= \left(\frac{6}{3}\right)^3$ $= (2)^3$ $= 8$

Why this step is taken: This is the final calculation phase. By systematically substituting the derived simplified forms and the calculated value of $|A|$, we arrive at the numerical answer.

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Tips and Common Mistakes to Avoid

• Order of Matrix ($n$): Always identify the order of the matrix first (here, $n=3$). Many formulas depend on $n$. A common mistake is using $n=2$ or $n=4$ incorrectly.
• Distinguishing Adjoint Properties: Be careful not to confuse $|\text{adj}X| = |X|^{n-1}$ with $|\text{adj}(\text{adj}X)| = |X|^{(n-1)^2}$. Understand their derivations.
• Scalar Multiplication of Determinants: Remember that $|kA| = k^n|A|$, not just $k|A|$. The scalar $k$ is raised to the power of the matrix's order $n$.
• Determinant Calculation: Practice calculating determinants of $3 \times 3$ matrices accurately. A small error in $|A|$ will propagate through the entire problem.
• Step-by-Step Approach: Break down complex problems into smaller, manageable steps. This reduces the chance of errors and makes the solution clearer.

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Summary and Key Takeaway

This problem is a classic example of how to efficiently use the properties of determinants and adjoints. The key takeaway is to:

1. Always calculate the fundamental determinant ($|A|$ in this case) first.
2. Systematically apply the relevant properties, such as $|\text{adj}X| = |X|^{n-1}$ and $|kX| = k^n|X|$, by carefully substituting $X$, $k$, and $n$.
3. Simplify the expression algebraically before substituting numerical values to avoid cumbersome calculations.

Mastering these properties is essential for solving a wide range of problems in matrices and determinants for JEE.

The final answer is $\boxed{\text{8}}$.