If the system of equation `x - 2y + 5z = 3`
`2x - y + z = 1` and `11x - 7y + pz = q` has infinitely many solution, then

Options

A. `p + q = 2`
B. `p + q = 10`
C. `p-q=2`
D. `p-q=5`

Answer

To determine the values of \( p \) and \( q \) for the system of equations to have infinitely many solutions, we will analyze the given equations step by step.

### Given Equations:
1. \( x - 2y + 5z = 3 \) (Equation 1)
2. \( 2x - y + z = 1 \) (Equation 2)
3. \( 11x - 7y + pz = q \) (Equation 3)

For the system to have infinitely many solutions, the determinant of the coefficients must be zero. This means that the equations are dependent.

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & -2 & 5 \\ 2 & -1 & 1 \\ 11 & -7 & p \end{bmatrix} \]

### Step 2: Calculate the Determinant
We need to calculate the determinant \( D \) of the matrix \( A \):
\[ D = \begin{vmatrix} 1 & -2 & 5 \\ 2 & -1 & 1 \\ 11 & -7 & p \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ D = 1 \cdot \begin{vmatrix} -1 & 1 \\ -7 & p \end{vmatrix} - (-2) \cdot \begin{vmatrix} 2 & 1 \\ 11 & p \end{vmatrix} + 5 \cdot \begin{vmatrix} 2 & -1 \\ 11 & -7 \end{vmatrix} \]

### Step 3: Calculate Each Minor
1. \( \begin{vmatrix} -1 & 1 \\ -7 & p \end{vmatrix} = (-1)p - (1)(-7) = -p + 7 \)

2. \( \begin{vmatrix} 2 & 1 \\ 11 & p \end{vmatrix} = (2p - 11) \)

3. \( \begin{vmatrix} 2 & -1 \\ 11 & -7 \end{vmatrix} = (2 \cdot -7) - (-1 \cdot 11) = -14 + 11 = -3 \)

### Step 4: Substitute Back into the Determinant
Now substituting back into the determinant:
\[ D = 1(-p + 7) + 2(2p - 11) + 5(-3) \]
\[ D = -p + 7 + 4p - 22 - 15 \]
\[ D = 3p - 30 \]

### Step 5: Set the Determinant to Zero
For the system to have infinitely many solutions:
\[ 3p - 30 = 0 \]
\[ 3p = 30 \implies p = 10 \]

### Step 6: Find \( q \) Using the Third Equation
Now we need to ensure that the third equation is also consistent with the first two. We set up the determinant of the augmented matrix:
\[ D' = \begin{vmatrix} 1 & -2 & 5 & 3 \\ 2 & -1 & 1 & 1 \\ 11 & -7 & 10 & q \end{vmatrix} \]

Calculating this determinant similarly, we find:
\[ D' = 1 \cdot \begin{vmatrix} -1 & 1 & 1 \\ -7 & 10 & q \end{vmatrix} - (-2) \cdot \begin{vmatrix} 2 & 1 & 1 \\ 11 & 10 & q \end{vmatrix} + 5 \cdot \begin{vmatrix} 2 & -1 & 1 \\ 11 & -7 & q \end{vmatrix} \]

### Step 7: Calculate Each Minor for \( D' \)
1. \( \begin{vmatrix} -1 & 1 & 1 \\ -7 & 10 & q \end{vmatrix} = -1(10q - 7) + 1(-7) = -10q + 7 - 7 = -10q \)

2. \( \begin{vmatrix} 2 & 1 & 1 \\ 11 & 10 & q \end{vmatrix} = 2(10 - q) - 1(11 - 10) = 20 - 2q - 11 = 9 - 2q \)

3. \( \begin{vmatrix} 2 & -1 & 1 \\ 11 & -7 & q \end{vmatrix} = 2(-7q + 11) - (-1)(11 - 2) = -14q + 22 + 11 - 2 = -14q + 31 \)

### Step 8: Substitute Back into \( D' \)
Substituting back:
\[ D' = 1(-10q) + 2(9 - 2q) + 5(-14q + 31) \]
\[ D' = -10q + 18 - 4q - 70q + 155 \]
\[ D' = -84q + 173 \]

### Step 9: Set \( D' \) to Zero
For consistency:
\[ -84q + 173 = 0 \implies 84q = 173 \implies q = \frac{173}{84} \]

### Final Step: Calculate \( p - q \) and \( p + q \)
1. \( p - q = 10 - \frac{173}{84} = \frac{840 - 173}{84} = \frac{667}{84} \)
2. \( p + q = 10 + \frac{173}{84} = \frac{840 + 173}{84} = \frac{1013}{84} \)

### Conclusion
Thus, the values of \( p \) and \( q \) are \( p = 10 \) and \( q = \frac{173}{84} \).