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. no solution
B. infinite number of solutions
C. exactly three solutions.
D. a unique solution

Explanation

To determine the number of solutions for the given system of linear equations, we will use the concept of determinants. The equations are:

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

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

We can represent the system of equations in matrix form as:

\[ \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 5 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 1 \end{bmatrix} \]

### Step 2: Calculate the Determinant \( D \)

The determinant \( D \) of the coefficient matrix is calculated as follows:

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

Using the determinant formula for a 3x3 matrix, we have:

\[ D = 1 \cdot (3 \cdot 2 - 1 \cdot 5) - 2 \cdot (2 \cdot 2 - 1 \cdot 3) + 1 \cdot (2 \cdot 5 - 3 \cdot 3) \]
\[ = 1 \cdot (6 - 5) - 2 \cdot (4 - 3) + 1 \cdot (10 - 9) \]
\[ = 1 \cdot 1 - 2 \cdot 1 + 1 \cdot 1 \]
\[ = 1 - 2 + 1 = 0 \]

### Step 3: Calculate Determinants \( D_1, D_2, D_3 \)

Next, we calculate \( D_1, D_2, D_3 \) by replacing the respective columns with the constant matrix.

#### Determinant \( D_1 \)

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

Calculating \( D_1 \):

\[ D_1 = 3 \cdot (3 \cdot 2 - 1 \cdot 5) - 2 \cdot (3 \cdot 2 - 1 \cdot 3) + 1 \cdot (3 \cdot 5 - 3 \cdot 1) \]
\[ = 3 \cdot (6 - 5) - 2 \cdot (6 - 3) + 1 \cdot (15 - 3) \]
\[ = 3 \cdot 1 - 2 \cdot 3 + 1 \cdot 12 \]
\[ = 3 - 6 + 12 = 9 \]

#### Determinant \( D_2 \)

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

Calculating \( D_2 \):

\[ D_2 = 1 \cdot (3 \cdot 2 - 1 \cdot 1) - 3 \cdot (2 \cdot 2 - 1 \cdot 3) + 1 \cdot (2 \cdot 1 - 3 \cdot 3) \]
\[ = 1 \cdot (6 - 1) - 3 \cdot (4 - 3) + 1 \cdot (2 - 9) \]
\[ = 1 \cdot 5 - 3 \cdot 1 + 1 \cdot (-7) \]
\[ = 5 - 3 - 7 = -5 \]

#### Determinant \( D_3 \)

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

Calculating \( D_3 \):

\[ D_3 = 1 \cdot (3 \cdot 1 - 3 \cdot 5) - 2 \cdot (2 \cdot 1 - 3 \cdot 3) + 3 \cdot (2 \cdot 5 - 3 \cdot 3) \]
\[ = 1 \cdot (3 - 15) - 2 \cdot (2 - 9) + 3 \cdot (10 - 9) \]
\[ = 1 \cdot (-12) - 2 \cdot (-7) + 3 \cdot 1 \]
\[ = -12 + 14 + 3 = 5 \]

### Step 4: Analyze the Determinants

Now we have:
- \( D = 0 \)
- \( D_1 = 9 \)
- \( D_2 = -5 \)
- \( D_3 = 5 \)

Since \( D = 0 \) and \( D_1 \neq 0 \), \( D_2 \neq 0 \), and \( D_3 \neq 0 \), this indicates that the system of equations has **no solution**.

### Conclusion

The system of linear equations has **no solution**.