`x_(1), x_(2), x_(3)` are three real numbers satisfying the system of equations
`x_(1)+3x_(2)+9x_(3)=27, x_(1)+5x_(2)+25x_(3)=125` and `x_(1)+7x_(2)+49x_(3)=343`, then which of the following options are correct

Options

A. number of divisor of `x_(1)+x_(3)` is 6
B. `(x_(1)+x_(2))/2` is a prime number
C. `x_(3)-x_(2)` is a prme number
D. `x_(1)+x_(2)+x_(3)` is square of an integer

Explanation

To solve the system of equations given by:

1. \( x_1 + 3x_2 + 9x_3 = 27 \) (Equation 1)
2. \( x_1 + 5x_2 + 25x_3 = 125 \) (Equation 2)
3. \( x_1 + 7x_2 + 49x_3 = 343 \) (Equation 3)

we will follow these steps:

### Step 1: Subtract Equation 1 from Equation 2
We will subtract Equation 1 from Equation 2 to eliminate \( x_1 \).

\[ (x_1 + 5x_2 + 25x_3) - (x_1 + 3x_2 + 9x_3) = 125 - 27 \]

This simplifies to:

\[ (5x_2 - 3x_2) + (25x_3 - 9x_3) = 98 \]

\[ 2x_2 + 16x_3 = 98 \]

Dividing through by 2 gives:

\[ x_2 + 8x_3 = 49 \quad \text{(Equation 4)} \]

### Step 2: Subtract Equation 2 from Equation 3
Next, we subtract Equation 2 from Equation 3:

\[ (x_1 + 7x_2 + 49x_3) - (x_1 + 5x_2 + 25x_3) = 343 - 125 \]

This simplifies to:

\[ (7x_2 - 5x_2) + (49x_3 - 25x_3) = 218 \]

\[ 2x_2 + 24x_3 = 218 \]

Dividing through by 2 gives:

\[ x_2 + 12x_3 = 109 \quad \text{(Equation 5)} \]

### Step 3: Solve for \( x_2 \) and \( x_3 \)
Now we have two new equations:

1. \( x_2 + 8x_3 = 49 \) (Equation 4)
2. \( x_2 + 12x_3 = 109 \) (Equation 5)

Subtract Equation 4 from Equation 5:

\[ (x_2 + 12x_3) - (x_2 + 8x_3) = 109 - 49 \]

This simplifies to:

\[ 4x_3 = 60 \]

Thus, we find:

\[ x_3 = 15 \]

### Step 4: Substitute \( x_3 \) back to find \( x_2 \)
Substituting \( x_3 = 15 \) into Equation 4:

\[ x_2 + 8(15) = 49 \]

\[ x_2 + 120 = 49 \]

\[ x_2 = 49 - 120 = -71 \]

### Step 5: Substitute \( x_2 \) and \( x_3 \) back to find \( x_1 \)
Now substitute \( x_2 = -71 \) and \( x_3 = 15 \) into Equation 1:

\[ x_1 + 3(-71) + 9(15) = 27 \]

\[ x_1 - 213 + 135 = 27 \]

\[ x_1 - 78 = 27 \]

\[ x_1 = 27 + 78 = 105 \]

### Summary of Values
We have:

- \( x_1 = 105 \)
- \( x_2 = -71 \)
- \( x_3 = 15 \)

### Step 6: Check the Options
Now we check the options given in the problem:

1. **Number of divisors of \( x_1 + x_3 \)**:
\[ x_1 + x_3 = 105 + 15 = 120 \]
The prime factorization of 120 is \( 2^3 \times 3^1 \times 5^1 \). The number of divisors is \( (3+1)(1+1)(1+1) = 4 \times 2 \times 2 = 16 \).

2. **Is \( \frac{x_1 + x_2}{2} \) a prime number?**:
\[ x_1 + x_2 = 105 - 71 = 34 \quad \Rightarrow \quad \frac{34}{2} = 17 \]
17 is a prime number.

3. **Is \( x_3 - x_2 \) a prime number?**:
\[ x_3 - x_2 = 15 - (-71) = 15 + 71 = 86 \]
86 is not a prime number.

4. **Is \( x_1 + x_2 + x_3 \) a square of an integer?**:
\[ x_1 + x_2 + x_3 = 105 - 71 + 15 = 49 \]
49 is \( 7^2 \), which is a perfect square.

### Conclusion
The correct options are 1, 2, and 4.