Let M be a square matix of order 3 whose elements are real number and `adj(adj M) = [(36,0,-4),(0,6,0),(0,3,6)]`, then the absolute value of `Tr(M)` is [Here, adj P denotes adjoint matrix of P and `T_(r) (P)` denotes trace of matrix P i.e., sum of all principal diagonal elements of matrix P]

Answer

To solve the problem step by step, we will analyze the given information and apply the necessary mathematical concepts.

### Step 1: Understand the given information
We are given that \( \text{adj}(\text{adj} M) = \begin{pmatrix} 36 & 0 & -4 \\ 0 & 6 & 0 \\ 0 & 3 & 6 \end{pmatrix} \).

### Step 2: Use the property of adjoint
We know that for a square matrix \( M \) of order \( n \):
\[ \text{adj}(\text{adj} M) = (\det M)^{n-1} M \]
For our case, \( n = 3 \), so:
\[ \text{adj}(\text{adj} M) = (\det M)^{3-1} M = (\det M)^2 M \]

### Step 3: Calculate the determinant of the adjoint matrix
To find \( \det(\text{adj}(\text{adj} M)) \), we can use the determinant of the given matrix:
\[ \det\left(\begin{pmatrix} 36 & 0 & -4 \\ 0 & 6 & 0 \\ 0 & 3 & 6 \end{pmatrix}\right) \]
Calculating this determinant:
\[ = 36 \cdot (6 \cdot 6 - 0 \cdot 3) - 0 + 0 = 36 \cdot 36 = 1296 \]

### Step 4: Set up the equation
From the property we used earlier:
\[ \det(\text{adj}(\text{adj} M)) = (\det M)^2 \cdot \det M = (\det M)^3 \]
Thus, we have:
\[ (\det M)^3 = 1296 \]

### Step 5: Solve for \( \det M \)
Taking the cube root of both sides:
\[ \det M = \sqrt[3]{1296} = 6 \text{ (since } 1296 = 6^4\text{)} \]
Thus, \( \det M = \pm 6 \).

### Step 6: Find the trace of \( M \)
Using the characteristic polynomial of \( M \):
\[ \lambda^3 - \text{Tr}(M) \lambda^2 + c_1 \lambda - \det M = 0 \]
Where \( c_1 \) is the sum of the products of the eigenvalues taken two at a time.

### Step 7: Relate the trace and determinant
Since we know \( \det M = \pm 6 \), we can assume possible eigenvalues \( \lambda_1, \lambda_2, \lambda_3 \) such that:
\[ \lambda_1 + \lambda_2 + \lambda_3 = \text{Tr}(M) \]
and
\[ \lambda_1 \lambda_2 \lambda_3 = \det M \]

### Step 8: Find possible values for the trace
Assuming \( \det M = 6 \):
Let’s take \( \lambda_1 = 2, \lambda_2 = 2, \lambda_3 = 2 \):
\[ \text{Tr}(M) = 2 + 2 + 2 = 6 \]
Assuming \( \det M = -6 \):
Let’s take \( \lambda_1 = 3, \lambda_2 = 2, \lambda_3 = -1 \):
\[ \text{Tr}(M) = 3 + 2 - 1 = 4 \]

### Final Step: Calculate the absolute value of the trace
Thus, the absolute value of \( \text{Tr}(M) \) can be:
\[ |\text{Tr}(M)| = |6| = 6 \quad \text{or} \quad |4| = 4 \]

### Conclusion
However, since we need the absolute value of the trace, we conclude:
\[ \text{The absolute value of } \text{Tr}(M) = 8. \]