For a unique value of `mu` & `lambda` , the system of equations given by
`{:(x+y+z=6),(x+2y+3z=14),(2x+5y+lambdaz=mu):}`
has infinitely many solutions , then `(mu-lambda)/4` is equal to

Answer

To solve the problem step by step, we need to analyze the given system of equations and determine the conditions under which it has infinitely many solutions.

### Given System of Equations:
1. \( x + y + z = 6 \) (Equation 1)
2. \( x + 2y + 3z = 14 \) (Equation 2)
3. \( 2x + 5y + \lambda z = \mu \) (Equation 3)

### Step 1: Form the Coefficient Matrix and Calculate the Determinant
To find the conditions for infinitely many solutions, we need to set the determinant \( D \) of the coefficient matrix equal to zero. The coefficient matrix is:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 5 & \lambda \end{bmatrix} \]

The determinant \( D \) can be calculated using the formula for the determinant of a 3x3 matrix:

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

For our matrix, we have:

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

Calculating each term:

- First term: \( 1 \cdot (2\lambda - 15) = 2\lambda - 15 \)
- Second term: \( -1 \cdot (\lambda - 6) = -\lambda + 6 \)
- Third term: \( 1 \cdot (5 - 4) = 1 \)

So, we have:

\[ D = (2\lambda - 15) - (\lambda - 6) + 1 = 2\lambda - 15 - \lambda + 6 + 1 = \lambda - 8 \]

### Step 2: Set the Determinant to Zero
For the system to have infinitely many solutions, we set \( D = 0 \):

\[ \lambda - 8 = 0 \]

Thus, we find:

\[ \lambda = 8 \]

### Step 3: Calculate the Determinant \( D_3 \)
Next, we need to find \( D_3 \) by replacing the third column of the coefficient matrix with the constants from the right-hand side of the equations:

\[ D_3 = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 14 \\ 2 & 5 & \mu \end{vmatrix} \]

Calculating \( D_3 \):

\[ D_3 = 1 \cdot (2\mu - 14 \cdot 5) - 1 \cdot (1 \cdot \mu - 14 \cdot 2) + 6 \cdot (1 \cdot 5 - 2 \cdot 1) \]

Calculating each term:

- First term: \( 1 \cdot (2\mu - 70) = 2\mu - 70 \)
- Second term: \( -1 \cdot (\mu - 28) = -\mu + 28 \)
- Third term: \( 6 \cdot (5 - 2) = 6 \cdot 3 = 18 \)

So, we have:

\[ D_3 = (2\mu - 70) - (\mu - 28) + 18 = 2\mu - 70 - \mu + 28 + 18 = \mu - 24 \]

### Step 4: Set \( D_3 = 0 \)
For the system to have infinitely many solutions, we also set \( D_3 = 0 \):

\[ \mu - 24 = 0 \]

Thus, we find:

\[ \mu = 24 \]

### Step 5: Calculate \( \frac{\mu - \lambda}{4} \)
Now we have \( \mu = 24 \) and \( \lambda = 8 \). We can calculate:

\[ \frac{\mu - \lambda}{4} = \frac{24 - 8}{4} = \frac{16}{4} = 4 \]

### Final Answer:
\[ \frac{\mu - \lambda}{4} = 4 \]