The system of equations x+2y +3z =4, 2x+3y+4z=5,3x+4y+5z=6 has

Options

A. Infinitely many solution
B. No solution
C. Unique solutions
D. none of these

Answer

To determine the type of solution for the given system of equations:

1. **Write down the equations:**
\[ \begin{align*} 1. & \quad x + 2y + 3z = 4 \\ 2. & \quad 2x + 3y + 4z = 5 \\ 3. & \quad 3x + 4y + 5z = 6 \end{align*} \]

2. **Identify the coefficients and constants:**
The coefficients of \(x\), \(y\), and \(z\) can be arranged in a matrix form:
\[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} \]
The constants on the right side are \(4\), \(5\), and \(6\).

3. **Calculate the determinant \(D\):**
The determinant \(D\) of the coefficient matrix can be calculated using the formula for a \(3 \times 3\) determinant:
\[ D = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} \]

Expanding this determinant:
\[ D = 1 \cdot (3 \cdot 5 - 4 \cdot 4) - 2 \cdot (2 \cdot 5 - 4 \cdot 3) + 3 \cdot (2 \cdot 4 - 3 \cdot 3) \]
\[ = 1 \cdot (15 - 16) - 2 \cdot (10 - 12) + 3 \cdot (8 - 9) \]
\[ = 1 \cdot (-1) - 2 \cdot (-2) + 3 \cdot (-1) \]
\[ = -1 + 4 - 3 = 0 \]

4. **Determine the values of \(D_x\), \(D_y\), and \(D_z\):**
- **Calculate \(D_x\):** Replace the first column with the constants:
\[ D_x = \begin{vmatrix} 4 & 2 & 3 \\ 5 & 3 & 4 \\ 6 & 4 & 5 \end{vmatrix} \]
Expanding this determinant:
\[ D_x = 4 \cdot (3 \cdot 5 - 4 \cdot 4) - 2 \cdot (5 \cdot 5 - 4 \cdot 6) + 3 \cdot (5 \cdot 4 - 3 \cdot 6) \]
\[ = 4 \cdot (15 - 16) - 2 \cdot (25 - 24) + 3 \cdot (20 - 18) \]
\[ = 4 \cdot (-1) - 2 \cdot 1 + 3 \cdot 2 \]
\[ = -4 - 2 + 6 = 0 \]

- **Calculate \(D_y\):** Replace the second column with the constants:
\[ D_y = \begin{vmatrix} 1 & 4 & 3 \\ 2 & 5 & 4 \\ 3 & 6 & 5 \end{vmatrix} \]
Expanding this determinant:
\[ D_y = 1 \cdot (5 \cdot 5 - 4 \cdot 6) - 4 \cdot (2 \cdot 5 - 4 \cdot 3) + 3 \cdot (2 \cdot 6 - 5 \cdot 3) \]
\[ = 1 \cdot (25 - 24) - 4 \cdot (10 - 12) + 3 \cdot (12 - 15) \]
\[ = 1 \cdot 1 - 4 \cdot (-2) + 3 \cdot (-3) \]
\[ = 1 + 8 - 9 = 0 \]

- **Calculate \(D_z\):** Replace the third column with the constants:
\[ D_z = \begin{vmatrix} 1 & 2 & 4 \\ 2 & 3 & 5 \\ 3 & 4 & 6 \end{vmatrix} \]
Expanding this determinant:
\[ D_z = 1 \cdot (3 \cdot 6 - 5 \cdot 4) - 2 \cdot (2 \cdot 6 - 5 \cdot 3) + 4 \cdot (2 \cdot 4 - 3 \cdot 3) \]
\[ = 1 \cdot (18 - 20) - 2 \cdot (12 - 15) + 4 \cdot (8 - 9) \]
\[ = 1 \cdot (-2) - 2 \cdot (-3) + 4 \cdot (-1) \]
\[ = -2 + 6 - 4 = 0 \]

5. **Conclusion:**
Since \(D = 0\), \(D_x = 0\), \(D_y = 0\), and \(D_z = 0\), we conclude that the system of equations has infinitely many solutions.

### Final Answer:
The system of equations has infinitely many solutions.