Let f be an even function defined on R. Suppose f is defined on [0,4] as follows:
`f(x) = {{:(3x, if, 0 le x lt 1),(4-x, if, 1 le x le 4):}`
and f satisfies the condition.
`f(x-2) = f{x + [(6x^(2) + 272)/(x^(2) + 2)]) AA x in R`
Where [x] = greatest integer `le x`. Then `f(3271) +f(-2052) + f(806)`=..............

Answer

To solve the problem step by step, we will analyze the function \( f \) and the given conditions.

### Step 1: Understand the Function Definition
The function \( f \) is defined as:
- \( f(x) = 3x \) for \( 0 \leq x < 1 \)
- \( f(x) = 4 - x \) for \( 1 \leq x \leq 4 \)

Since \( f \) is an even function, we have:
- \( f(-x) = f(x) \) for all \( x \in \mathbb{R} \)

### Step 2: Analyze the Condition
The condition given is:
\[ f(x-2) = f\left(x + \frac{6x^2 + 272}{x^2 + 2}\right) \]
This implies that the function has a periodic behavior.

### Step 3: Calculate the Greatest Integer Function
We need to analyze the term:
\[ \frac{6x^2 + 272}{x^2 + 2} \]
Let's simplify it:
\[ \frac{6x^2 + 272}{x^2 + 2} = 6 + \frac{260}{x^2 + 2} \]
Since \( x^2 + 2 \) is always positive, \( \frac{260}{x^2 + 2} \) will be a positive number less than \( 6 \) for large \( x \). Thus, the greatest integer function \( \left\lfloor \frac{6x^2 + 272}{x^2 + 2} \right\rfloor \) will yield \( 6 \).

### Step 4: Use the Condition to Find Periodicity
From the condition:
\[ f(x-2) = f(x) + 6 \]
If we replace \( x \) with \( x + 2 \):
\[ f(x) = f(x + 2) + 6 \]
This indicates that the function is periodic with a period of \( 8 \) (since \( f(x + 8) = f(x) \)).

### Step 5: Evaluate \( f(3271) \)
To find \( f(3271) \):
1. Calculate \( 3271 \mod 8 \):
\[ 3271 \div 8 = 408 \quad \text{(remainder 7)} \]
So, \( 3271 \equiv 7 \mod 8 \).
2. Thus, \( f(3271) = f(7) \).
3. Since \( 7 > 4 \), we can find \( f(7) \) using the periodicity:
\[ f(7) = f(7 - 8) = f(-1) = f(1) = 4 - 1 = 3 \]

### Step 6: Evaluate \( f(-2052) \)
1. Calculate \( -2052 \mod 8 \):
\[ -2052 \div 8 = -257 \quad \text{(remainder 4)} \]
So, \( -2052 \equiv 4 \mod 8 \).
2. Thus, \( f(-2052) = f(4) \).
3. From the function definition:
\[ f(4) = 4 - 4 = 0 \]

### Step 7: Evaluate \( f(806) \)
1. Calculate \( 806 \mod 8 \):
\[ 806 \div 8 = 100 \quad \text{(remainder 6)} \]
So, \( 806 \equiv 6 \mod 8 \).
2. Thus, \( f(806) = f(6) \).
3. Since \( 6 > 4 \):
\[ f(6) = f(-2) = f(2) \quad \text{(because \( f \) is even)} \]
4. From the function definition:
\[ f(2) = 4 - 2 = 2 \]

### Step 8: Calculate the Final Result
Now, we can sum the values:
\[ f(3271) + f(-2052) + f(806) = 3 + 0 + 2 = 5 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{5} \]