Step 1: Understanding the Concept:
This question requires knowledge of determinant properties, specifically those involving scalar multiplication and the adjoint of a matrix.
Property 1: For any square matrix $M$ of order $n$ and scalar $k$, $|kM| = k^n |M|$.
Property 2: For any square matrix $A$ of order $n$, the determinant of its adjoint is given by $|adj A| = |A|^{n-1}$.
Combining these allows us to convert an equation involving the adjoint into a simple algebraic equation for $|A|$.
Step 2: Key Formula or Approach:
1. Use the property $|k \cdot M| = k^n \cdot |M|$ to handle the constant 2.
2. Use the property $|adj A| = |A|^{n-1}$ for $n=3$.
3. Solve the resulting quadratic or higher-degree equation.
Step 3: Detailed Explanation:
Given $n = 3$ (order of matrix $A$).
The equation is $|2(adj A)| = 288$.
According to Property 1, when a scalar 2 is inside a determinant of a $3 \times 3$ matrix, it comes out as $2^3$:
\[ 2^3 |adj A| = 288 \]
\[ 8 |adj A| = 288 \]
Dividing both sides by 8:
\[ |adj A| = \frac{288}{8} = 36 \]
Now, apply Property 2 where $n=3$:
\[ |adj A| = |A|^{3-1} = |A|^2 \]
Substituting the value we found:
\[ |A|^2 = 36 \]
Taking the square root on both sides:
\[ |A| = \pm \sqrt{36} = \pm 6 \]
This gives two possible values, $+6$ and $-6$, which is represented in option (D).
Step 4: Final Answer:
The possible values of the determinant $|A|$ are $\pm 6$.