For what values of p and q the system of equations `2x+py+6z=8, x+2y+qz=5, x+y+3z=4` has a unique solution. (a).`p = 2, q ne3` (b).`p ne 2, q ne 3` (c).`p ne 2, q = 3` (d).`p =2, q=3`

Options

A. `p = 2, q in 3`
B. `p in 2, qin 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

\[ \begin{align*} 1. & \quad 2x + py + 6z = 8 \\ 2. & \quad x + 2y + qz = 5 \\ 3. & \quad x + y + 3z = 4 \end{align*} \]

has a unique solution, we need to analyze the coefficient matrix of the system. The condition for a system of linear equations to have a unique solution is that the determinant of the coefficient matrix must be non-zero.

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) for the given system is:

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

### Step 2: Calculate the Determinant
We need to calculate the determinant of matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix, we expand along the first row:

\[ \text{det}(A) = 2 \begin{vmatrix} 2 & q \\ 1 & 3 \end{vmatrix} - p \begin{vmatrix} 1 & q \\ 1 & 3 \end{vmatrix} + 6 \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

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

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

Substituting these back into the determinant expression:

\[ \text{det}(A) = 2(6 - q) - p(3 - q) + 6(-1) \]

### Step 3: Simplify the Determinant
Now simplifying the expression:

\[ \text{det}(A) = 12 - 2q - 3p + pq - 6 \]
\[ = pq - 3p - 2q + 6 \]

### Step 4: Set the Determinant Not Equal to Zero
For the system to have a unique solution, we require:

\[ pq - 3p - 2q + 6 \neq 0 \]

### Step 5: Factor the Expression
We can rearrange this as:

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

### Step 6: Determine Conditions for Non-Zero
For the product \( (p - 2)(q - 3) \) to be non-zero, both factors must be non-zero:

1. \( p - 2 \neq 0 \) implies \( p \neq 2 \)
2. \( q - 3 \neq 0 \) implies \( q \neq 3 \)

### Conclusion
Thus, the values of \( p \) and \( q \) for which the system has a unique solution are:

**Answer:** \( p \neq 2 \) and \( q \neq 3 \).

The correct option is (b) \( p \neq 2, q \neq 3 \).