Consider the system of linear equations:
`x_(1) + 2x_(2) + x_(3) = 3`
`2x_(1) + 3x_(2) + x_(3) = 3`
`3x_(1) + 5x_(2) + 2x_(3) = 1`
The system has

Options

A. Infinite number of solutions
B. Exactly 3 solutions
C. A unique solution
D. No solution

Answer

To solve the given system of linear equations, we will use determinants to analyze the system. The equations are:

1. \( x_1 + 2x_2 + x_3 = 3 \)
2. \( 2x_1 + 3x_2 + x_3 = 3 \)
3. \( 3x_1 + 5x_2 + 2x_3 = 1 \)

We want to determine if the system has a unique solution, no solution, or infinitely many solutions. We will use Cramer's rule and the concept of determinants.

### Step 1: Form the Coefficient Matrix and the Constant Matrix

The coefficient matrix \( A \) and the constant matrix \( B \) can be formed as follows:

\[ A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 3 \\ 3 \\ 1 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix \( A \)

We will calculate the determinant \( D \) of the matrix \( A \):

\[ D = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

Calculating \( D \):

\[ D = 1 \cdot (3 \cdot 2 - 1 \cdot 5) - 2 \cdot (2 \cdot 2 - 1 \cdot 3) + 1 \cdot (2 \cdot 5 - 3 \cdot 3) \]

Calculating each term:

1. \( 3 \cdot 2 - 1 \cdot 5 = 6 - 5 = 1 \)
2. \( 2 \cdot 2 - 1 \cdot 3 = 4 - 3 = 1 \)
3. \( 2 \cdot 5 - 3 \cdot 3 = 10 - 9 = 1 \)

Putting it all together:

\[ D = 1 \cdot 1 - 2 \cdot 1 + 1 \cdot 1 = 1 - 2 + 1 = 0 \]

### Step 3: Analyze the Determinant

Since \( D = 0 \), we need to check the determinants of the matrices formed by replacing columns of \( A \) with the constant matrix \( B \).

### Step 4: Calculate \( D_1 \), \( D_2 \), and \( D_3 \)

#### Calculate \( D_1 \)

Replace the first column of \( A \) with \( B \):

\[ D_1 = \begin{vmatrix} 3 & 2 & 1 \\ 3 & 3 & 1 \\ 1 & 5 & 2 \end{vmatrix} \]

Calculating \( D_1 \):

\[ D_1 = 3(3 \cdot 2 - 1 \cdot 5) - 2(3 \cdot 2 - 1 \cdot 1) + 1(3 \cdot 5 - 3 \cdot 1) \]

Calculating each term:

1. \( 3 \cdot 2 - 1 \cdot 5 = 6 - 5 = 1 \)
2. \( 3 \cdot 2 - 1 \cdot 1 = 6 - 1 = 5 \)
3. \( 3 \cdot 5 - 3 \cdot 1 = 15 - 3 = 12 \)

Putting it all together:

\[ D_1 = 3 \cdot 1 - 2 \cdot 5 + 1 \cdot 12 = 3 - 10 + 12 = 5 \]

#### Calculate \( D_2 \)

Replace the second column of \( A \) with \( B \):

\[ D_2 = \begin{vmatrix} 1 & 3 & 1 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix} \]

Calculating \( D_2 \):

\[ D_2 = 1(3 \cdot 2 - 1 \cdot 1) - 3(2 \cdot 2 - 1 \cdot 3) + 1(2 \cdot 1 - 3 \cdot 3) \]

Calculating each term:

1. \( 3 \cdot 2 - 1 \cdot 1 = 6 - 1 = 5 \)
2. \( 2 \cdot 2 - 1 \cdot 3 = 4 - 3 = 1 \)
3. \( 2 \cdot 1 - 3 \cdot 3 = 2 - 9 = -7 \)

Putting it all together:

\[ D_2 = 1 \cdot 5 - 3 \cdot 1 + 1 \cdot (-7) = 5 - 3 - 7 = -5 \]

#### Calculate \( D_3 \)

Replace the third column of \( A \) with \( B \):

\[ D_3 = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 3 \\ 3 & 5 & 1 \end{vmatrix} \]

Calculating \( D_3 \):

\[ D_3 = 1(3 \cdot 1 - 3 \cdot 5) - 2(2 \cdot 1 - 3 \cdot 3) + 3(2 \cdot 5 - 3 \cdot 3) \]

Calculating each term:

1. \( 3 \cdot 1 - 3 \cdot 5 = 3 - 15 = -12 \)
2. \( 2 \cdot 1 - 3 \cdot 3 = 2 - 9 = -7 \)
3. \( 2 \cdot 5 - 3 \cdot 3 = 10 - 9 = 1 \)

Putting it all together:

\[ D_3 = 1 \cdot (-12) - 2 \cdot (-7) + 3 \cdot 1 = -12 + 14 + 3 = 5 \]

### Step 5: Conclusion

We have:
- \( D = 0 \)
- \( D_1 = 5 \) (non-zero)
- \( D_2 = -5 \) (non-zero)
- \( D_3 = 5 \) (non-zero)

Since \( D = 0 \) and at least one of \( D_1, D_2, D_3 \) is non-zero, the system of equations has no solution.

### Final Answer

The system has **no solution**.