For what values of p and q the system of equations `2x+py+6z=8, x+2y+qz=5, x+y+3z=4` has i no solution ii a unique solution iii in finitely many solutions.

Options

A. `p=2, q ne 3`.
B. `p ne 2, q ne 3`
C. `p ne 2, q = 3`
D. `p =2, q=3`

Explanation

To determine the values of \( p \) and \( q \) for which the system of equations has no solution, a unique solution, or infinitely many solutions, we will analyze the given equations:

1. \( 2x + py + 6z = 8 \) (Equation 1)
2. \( x + 2y + qz = 5 \) (Equation 2)
3. \( x + y + 3z = 4 \) (Equation 3)

We will represent this system in matrix form and use determinants to analyze the conditions for the solutions.

### Step 1: Write the augmented matrix
The augmented matrix for the system of equations is:

\[ \begin{bmatrix} 2 & p & 6 & | & 8 \\ 1 & 2 & q & | & 5 \\ 1 & 1 & 3 & | & 4 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
The coefficient matrix is:

\[ A = \begin{bmatrix} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{bmatrix} \]

To find the determinant \( \Delta \) of matrix \( A \):

\[ \Delta = \begin{vmatrix} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \Delta = 2 \begin{vmatrix} 2 & q \\ 1 & 3 \end{vmatrix} - p \begin{vmatrix} 1 & q \\ 1 & 3 \end{vmatrix} + 6 \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating the minors:

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

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

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

Substituting back into the determinant formula:

\[ \Delta = 2(6 - q) - p(3 - q) + 6(-1) \]

Simplifying:

\[ \Delta = 12 - 2q - 3p + pq - 6 = pq - 3p - 2q + 6 \]

### Step 3: Conditions for solutions
1. **No Solution**: The system has no solution if \( \Delta = 0 \) and at least one of the other determinants (for \( \Delta_y \) or \( \Delta_z \)) is non-zero.

2. **Unique Solution**: The system has a unique solution if \( \Delta \neq 0 \).

3. **Infinitely Many Solutions**: The system has infinitely many solutions if \( \Delta = 0 \) and all other determinants are also zero.

### Step 4: Finding conditions
- For **no solution**:
Set \( \Delta = 0 \):
\[ pq - 3p - 2q + 6 = 0 \]
This means \( p \) and \( q \) must satisfy this equation, and at least one of the other determinants must be non-zero.

- For **unique solution**:
We need \( pq - 3p - 2q + 6 \neq 0 \).

- For **infinitely many solutions**:
Set \( \Delta = 0 \) and check if the other determinants also equal zero.

### Conclusion
To summarize, the conditions for \( p \) and \( q \) are:

- **No Solution**: \( pq - 3p - 2q + 6 = 0 \) and at least one of the other determinants is non-zero.
- **Unique Solution**: \( pq - 3p - 2q + 6 \neq 0 \).
- **Infinitely Many Solutions**: \( pq - 3p - 2q + 6 = 0 \) and all other determinants are zero.