If `3^n` is a factor of the determinant `|{:(1,1,1),(.^nC_1,.^(n+3)C_1,.^(n+6)C_1),(.^nC_2, .^(n+3)C_2, .^(n+6)C_2):}|` then the maximum value of n is ……..

Answer

To solve the problem, we need to find the maximum value of \( n \) such that \( 3^n \) is a factor of the determinant

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ \binom{n}{1} & \binom{n+3}{1} & \binom{n+6}{1} \\ \binom{n}{2} & \binom{n+3}{2} & \binom{n+6}{2} \end{vmatrix} \]

### Step 1: Calculate the Binomial Coefficients
The binomial coefficients are calculated as follows:
- \( \binom{n}{1} = n \)
- \( \binom{n+3}{1} = n + 3 \)
- \( \binom{n+6}{1} = n + 6 \)
- \( \binom{n}{2} = \frac{n(n-1)}{2} \)
- \( \binom{n+3}{2} = \frac{(n+3)(n+2)}{2} \)
- \( \binom{n+6}{2} = \frac{(n+6)(n+5)}{2} \)

### Step 2: Substitute the Values into the Determinant
Substituting the values of the binomial coefficients into the determinant gives:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ n & n+3 & n+6 \\ \frac{n(n-1)}{2} & \frac{(n+3)(n+2)}{2} & \frac{(n+6)(n+5)}{2} \end{vmatrix} \]

### Step 3: Factor Out Common Terms
We can factor out \( \frac{1}{2} \) from the third row:

\[ D = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ n & n+3 & n+6 \\ n(n-1) & (n+3)(n+2) & (n+6)(n+5) \end{vmatrix} \]

### Step 4: Apply Row Operations
We can perform row operations to simplify the determinant. Subtract the first row from the second and third rows:

\[ D = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 0 & 3 & 6 \\ 0 & (n+3)(n+2) - n(n-1) & (n+6)(n+5) - n(n-1) \end{vmatrix} \]

### Step 5: Simplify the Determinant
Now we need to calculate the elements in the third row:

1. For the second column:
\[ (n+3)(n+2) - n(n-1) = n^2 + 5n + 6 - (n^2 - n) = 6n + 6 \]

2. For the third column:
\[ (n+6)(n+5) - n(n-1) = n^2 + 11n + 30 - (n^2 - n) = 12n + 30 \]

Thus, the determinant simplifies to:

\[ D = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 0 & 3 & 6 \\ 0 & 6n + 6 & 12n + 30 \end{vmatrix} \]

### Step 6: Calculate the Determinant
Calculating the determinant, we have:

\[ D = \frac{1}{2} \cdot 1 \cdot \begin{vmatrix} 3 & 6 \\ 6n + 6 & 12n + 30 \end{vmatrix} \]

Calculating this 2x2 determinant:

\[ = \frac{1}{2} \cdot (3(12n + 30) - 6(6n + 6)) = \frac{1}{2} \cdot (36n + 90 - 36n - 36) = \frac{1}{2} \cdot 54 = 27 \]

### Step 7: Determine the Power of 3
Now we need to find the maximum \( n \) such that \( 3^n \) divides \( 27 \):

\[ 27 = 3^3 \]

Thus, the maximum value of \( n \) such that \( 3^n \) divides \( 27 \) is:

\[ \boxed{3} \]