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 given problem, we need to analyze the system of equations and determine the conditions under which it has infinitely many solutions. The equations are:

1. \( x + y + z = 6 \) (Equation 1)
2. \( x + 2y + 3z = 14 \) (Equation 2)
3. \( 2x + 5y + \lambda z = \mu \) (Equation 3)

### Step 1: Formulate the Determinant
For the system to have infinitely many solutions, the determinant of the coefficients must be zero. We denote the determinant as \( D \):

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

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

\[ D = 1 \cdot \begin{vmatrix} 2 & 3 \\ 5 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 3 \\ 2 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & 3 \\ 5 & \lambda \end{vmatrix} = 2\lambda - 15 \)

2. \( \begin{vmatrix} 1 & 3 \\ 2 & \lambda \end{vmatrix} = \lambda - 6 \)

3. \( \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = 5 - 4 = 1 \)

Putting these into the determinant \( D \):

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

### Step 3: Set the Determinant to Zero
For the system to have infinitely many solutions, we set \( D = 0 \):

\[ \lambda - 8 = 0 \implies \lambda = 8 \]

### Step 4: Calculate \( D_3 \)
Next, we need to find \( D_3 \) to determine \( \mu \):

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

Calculating \( D_3 \):

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

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 2 & 14 \\ 5 & \mu \end{vmatrix} = 2\mu - 70 \)

2. \( \begin{vmatrix} 1 & 14 \\ 2 & \mu \end{vmatrix} = \mu - 28 \)

3. \( \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = 1 \)

Putting these into \( D_3 \):

\[ D_3 = 1(2\mu - 70) - 1(\mu - 28) + 6(1) \]
\[ D_3 = 2\mu - 70 - \mu + 28 + 6 \]
\[ D_3 = \mu - 36 \]

### Step 5: Set \( D_3 \) to Zero
For the system to have infinitely many solutions, we set \( D_3 = 0 \):

\[ \mu - 36 = 0 \implies \mu = 36 \]

### Step 6: Calculate \( \frac{\mu - \lambda}{4} \)
Now, we need to find \( \frac{\mu - \lambda}{4} \):

\[ \frac{\mu - \lambda}{4} = \frac{36 - 8}{4} = \frac{28}{4} = 7 \]

### Final Answer
Thus, the value of \( \frac{\mu - \lambda}{4} \) is \( 7 \).

---