The greatest value of n for which the determinant
`Delta = |(1,1,1),(.^(n)C_(1),.^(n+3)C_(1),.^(n+6)C_(1)),(.^(n)C_(2),.^(n+3)C_(2),.^(n+6)C_(2))|` is divisible by `3^(n)`, is

Answer

To solve the problem, we need to find the greatest value of \( n \) for which the determinant

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

is divisible by \( 3^n \).

### Step 1: Write the Binomial Coefficients
First, we can express the binomial coefficients in terms of factorials:

- \(\binom{n}{1} = n\)
- \(\binom{n+3}{1} = n + 3\)
- \(\binom{n+6}{1} = n + 6\)
- \(\binom{n}{2} = \frac{n(n-1)}{2}\)
- \(\binom{n+3}{2} = \frac{(n+3)(n+2)}{2}\)
- \(\binom{n+6}{2} = \frac{(n+6)(n+5)}{2}\)

Substituting these values into the determinant, we have:

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

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

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

### Step 3: Perform Row Operations
Next, we can simplify the determinant by performing row operations. We can subtract the first column from the second and third columns:

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

Calculating the new entries in the third row:

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

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

Thus, we have:

\[ \Delta = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ n & 3 & 6 \\ n(n-1) & 6n + 6 & 12n + 30 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now we can compute the determinant:

\[ \Delta = \frac{1}{2} \left( 1 \cdot \begin{vmatrix} n & 3 \\ n(n-1) & 6n + 6 \end{vmatrix} - 1 \cdot \begin{vmatrix} n & 6 \\ n(n-1) & 12n + 30 \end{vmatrix} + 1 \cdot \begin{vmatrix} n & 3 \\ n & 6 \end{vmatrix} \right) \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} n & 3 \\ n(n-1) & 6n + 6 \end{vmatrix} = n(6n + 6) - 3n(n-1) = 6n^2 + 6n - 3n^2 + 3n = 3n^2 + 9n \)

2. \( \begin{vmatrix} n & 6 \\ n(n-1) & 12n + 30 \end{vmatrix} = n(12n + 30) - 6n(n-1) = 12n^2 + 30n - 6n^2 + 6n = 6n^2 + 36n \)

3. \( \begin{vmatrix} n & 3 \\ n & 6 \end{vmatrix} = n \cdot 6 - n \cdot 3 = 3n \)

Putting it all together:

\[ \Delta = \frac{1}{2} \left( (3n^2 + 9n) - (6n^2 + 36n) + 3n \right) = \frac{1}{2} \left( -3n^2 - 24n \right) = -\frac{3}{2}(n^2 + 8n) \]

### Step 5: Check Divisibility by \( 3^n \)
Now we need to check when \( -\frac{3}{2}(n^2 + 8n) \) is divisible by \( 3^n \).

The expression \( -\frac{3}{2}(n^2 + 8n) \) has a factor of \( 3 \). For it to be divisible by \( 3^n \), we need \( n^2 + 8n \) to be divisible by \( 3^{n-1} \).

### Step 6: Find the Maximum \( n \)
We can analyze the expression \( n^2 + 8n \):

- For \( n = 0 \): \( 0^2 + 8 \cdot 0 = 0 \) (divisible by \( 1 \))
- For \( n = 1 \): \( 1^2 + 8 \cdot 1 = 9 \) (divisible by \( 3 \))
- For \( n = 2 \): \( 2^2 + 8 \cdot 2 = 20 \) (not divisible by \( 9 \))
- For \( n = 3 \): \( 3^2 + 8 \cdot 3 = 51 \) (not divisible by \( 27 \))

Thus, the maximum \( n \) for which \( \Delta \) is divisible by \( 3^n \) is \( n = 3 \).

### Final Answer
The greatest value of \( n \) for which the determinant \( \Delta \) is divisible by \( 3^n \) is:

\[ \boxed{3} \]