`4x + 3y + 2z =1`
`x-y + 3z = 4`
`2x + 5y - 4z = 6` has n solution than n =

Explanation

To determine the number of solutions \( n \) for the given system of equations, we will analyze the equations step by step using determinants.

### Given Equations:
1. \( 4x + 3y + 2z = 1 \) (Equation 1)
2. \( x - y + 3z = 4 \) (Equation 2)
3. \( 2x + 5y - 4z = 6 \) (Equation 3)

### Step 1: Formulate the Coefficient Matrix
The coefficient matrix \( A \) for the system of equations is:
\[ A = \begin{bmatrix} 4 & 3 & 2 \\ 1 & -1 & 3 \\ 2 & 5 & -4 \end{bmatrix} \]

### Step 2: Calculate the Determinant of Matrix \( A \)
We will calculate the determinant \( |A| \) using the formula for the determinant of a 3x3 matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Where:
- \( a = 4, b = 3, c = 2 \)
- \( d = 1, e = -1, f = 3 \)
- \( g = 2, h = 5, i = -4 \)

Calculating the determinant:
\[ |A| = 4((-1)(-4) - (3)(5)) - 3((1)(-4) - (3)(2)) + 2((1)(5) - (-1)(2)) \]
\[ = 4(4 - 15) - 3(-4 - 6) + 2(5 + 2) \]
\[ = 4(-11) - 3(-10) + 2(7) \]
\[ = -44 + 30 + 14 \]
\[ = -44 + 44 = 0 \]

### Step 3: Analyze the Determinant
Since \( |A| = 0 \), we need to check for the conditions of the system of equations:
- If \( |A| = 0 \), the system may have either no solutions or infinitely many solutions.

### Step 4: Calculate the Determinants for the Conditions
We need to calculate \( \Delta_1, \Delta_2, \Delta_3 \) to determine the nature of the solutions.

#### Determinant \( \Delta_1 \):
Replace the first column of \( A \) with the constants from the right-hand side of the equations:
\[ \Delta_1 = \begin{vmatrix} 1 & 3 & 2 \\ 4 & -1 & 3 \\ 6 & 5 & -4 \end{vmatrix} \]
Calculating \( \Delta_1 \):
\[ = 1((-1)(-4) - (3)(5)) - 3((4)(-4) - (3)(6)) + 2((4)(5) - (-1)(6)) \]
\[ = 1(4 - 15) - 3(-16 - 18) + 2(20 + 6) \]
\[ = 1(-11) + 3(34) + 2(26) \]
\[ = -11 + 102 + 52 = 143 \]

#### Determinant \( \Delta_2 \):
Replace the second column of \( A \) with the constants:
\[ \Delta_2 = \begin{vmatrix} 4 & 1 & 2 \\ 1 & 4 & 3 \\ 2 & 6 & -4 \end{vmatrix} \]
Calculating \( \Delta_2 \):
\[ = 4(4 \cdot -4 - 3 \cdot 6) - 1(1 \cdot -4 - 3 \cdot 2) + 2(1 \cdot 6 - 4 \cdot 2) \]
\[ = 4(-16 - 18) - 1(-4 - 6) + 2(6 - 8) \]
\[ = 4(-34) + 10 - 4 = -136 + 10 - 4 = -130 \]

#### Determinant \( \Delta_3 \):
Replace the third column of \( A \) with the constants:
\[ \Delta_3 = \begin{vmatrix} 4 & 3 & 1 \\ 1 & -1 & 4 \\ 2 & 5 & 6 \end{vmatrix} \]
Calculating \( \Delta_3 \):
\[ = 4((-1)(6) - (4)(5)) - 3((1)(6) - (4)(2)) + 1((1)(5) - (-1)(2)) \]
\[ = 4(-6 - 20) - 3(6 - 8) + 1(5 + 2) \]
\[ = 4(-26) - 3(-2) + 7 = -104 + 6 + 7 = -91 \]

### Step 5: Analyze the Results
- \( \Delta_1 \neq 0 \)
- \( \Delta_2 \neq 0 \)
- \( \Delta_3 \neq 0 \)

Since \( \Delta_1 \neq 0 \) and \( |A| = 0 \), the system of equations has no solutions.

### Conclusion
Thus, the value of \( n \) is:
\[ n = 0 \]