If the system of equations `x+y+z=6, x+2y+lambdaz=10` and `x+2y+3z=mu` has infinite solutions, then the value of `lambda+2mu` is equal to

Options

A. 20
B. 22
C. 23
D. 25

Answer

To solve the problem step by step, we need to analyze the given system of equations and find the conditions under which they have infinite solutions.

### Given Equations:
1. \( x + y + z = 6 \) (Equation 1)
2. \( x + 2y + \lambda z = 10 \) (Equation 2)
3. \( x + 2y + 3z = \mu \) (Equation 3)

### Step 1: Set Up the Determinant for Infinite Solutions
For the system to have infinite solutions, the determinant of the coefficients of the variables must be zero. This determinant is given by:

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

### Step 2: Calculate the Determinant
Calculating the determinant:

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

Calculating the 2x2 determinants:

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

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

3. \( \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = (1 \cdot 2) - (1 \cdot 2) = 0 \)

Putting these back into the determinant:

\[ \Delta = 1(6 - 2\lambda) - 1(3 - \lambda) + 1(0) \]
\[ \Delta = 6 - 2\lambda - 3 + \lambda = 3 - \lambda \]

### Step 3: Set the Determinant to Zero
For infinite solutions, we set the determinant to zero:

\[ 3 - \lambda = 0 \implies \lambda = 3 \]

### Step 4: Find the Value of \(\Delta_1\)
Next, we need to find \(\Delta_1\) by replacing the first column of the determinant with the right-hand side values:

\[ \Delta_1 = \begin{vmatrix} 6 & 1 & 1 \\ 10 & 1 & \lambda \\ \mu & 1 & 3 \end{vmatrix} \]

Calculating this determinant:

\[ \Delta_1 = 6 \cdot \begin{vmatrix} 1 & \lambda \\ 1 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 10 & \lambda \\ \mu & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 10 & 1 \\ \mu & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

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

2. \( \begin{vmatrix} 10 & \lambda \\ \mu & 3 \end{vmatrix} = 10 \cdot 3 - \lambda \cdot \mu = 30 - \lambda \mu \)

3. \( \begin{vmatrix} 10 & 1 \\ \mu & 1 \end{vmatrix} = 10 \cdot 1 - 1 \cdot \mu = 10 - \mu \)

Putting these back into \(\Delta_1\):

\[ \Delta_1 = 6(3 - \lambda) - (30 - \lambda \mu) + (10 - \mu) \]
Substituting \(\lambda = 3\):

\[ \Delta_1 = 6(3 - 3) - (30 - 3\mu) + (10 - \mu) \]
\[ \Delta_1 = 0 - 30 + 3\mu + 10 - \mu = 2\mu - 20 \]

### Step 5: Set \(\Delta_1\) to Zero
For infinite solutions:

\[ 2\mu - 20 = 0 \implies 2\mu = 20 \implies \mu = 10 \]

### Step 6: Calculate \(\lambda + 2\mu\)
Now we have \(\lambda = 3\) and \(\mu = 10\):

\[ \lambda + 2\mu = 3 + 2(10) = 3 + 20 = 23 \]

### Final Answer
Thus, the value of \(\lambda + 2\mu\) is:

\[ \boxed{23} \]