If $A = \begin{bmatrix} a & 1 & -1 \\ 0 & b & 4 \\ 4 & 4 & c \end{bmatrix}$ and $abc = 12$, $b = 4a$, then the value of $|A(adjA)|$ is:

For any $n \times n$ matrix $A$:

$A \cdot adj(A) = |A| \cdot I$

where $I$ is the identity matrix and $|A|$ is the determinant of $A$.

Taking the determinant of both sides:

$|A \cdot adj(A)| = ||A| \cdot I|$

For a $3 \times 3$ matrix:

$|A \cdot adj(A)| = |A|^3$

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Given: $A = \begin{bmatrix} a & 1 & -1 \\ 0 & b & 4 \\ 4 & 4 & c \end{bmatrix}$

Using the first row expansion:

$|A| = a \begin{vmatrix} b & 4 \\ 4 & c \end{vmatrix} - 1 \begin{vmatrix} 0 & 4 \\ 4 & c \end{vmatrix} + (-1) \begin{vmatrix} 0 & b \\ 4 & 4 \end{vmatrix}$

Calculating each $2 \times 2$ determinant:

$\begin{vmatrix} b & 4 \\ 4 & c \end{vmatrix} = bc - 16$

$\begin{vmatrix} 0 & 4 \\ 4 & c \end{vmatrix} = 0 - 16 = -16$

$\begin{vmatrix} 0 & b \\ 4 & 4 \end{vmatrix} = 0 - 4b = -4b$

Therefore:

$|A| = a(bc - 16) - 1(-16) + (-1)(-4b)$

$|A| = abc - 16a + 16 + 4b$

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Given conditions: $abc = 12$ and $b = 4a$

Substituting these values:

$|A| = 12 - 16a + 16 + 4(4a)$

$|A| = 12 - 16a + 16 + 16a$

$|A| = 28$

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$|A(adjA)| = |A|^3$

$|A(adjA)| = 28^3$

Therefore, the value of $|A(adjA)|$ is $28^3$.