If the system of equations 2x+ay+6z=8, x+2y+z=5, 2x+ay+3z=4 has a unique solution then 'a' cannot be equal to :

Options

A. 2
B. 3
C. 4
D. 5

Answer

To determine the value of 'a' for which the system of equations has a unique solution, we need to find the determinant of the coefficient matrix and set it not equal to zero. The given equations are:

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

### Step 1: Write the coefficient matrix
The coefficient matrix \(A\) for the system of equations is:

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

### Step 2: Calculate the determinant of the matrix
To find the determinant of matrix \(A\), we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

- \(a = 2\), \(b = a\), \(c = 6\)
- \(d = 1\), \(e = 2\), \(f = 1\)
- \(g = 2\), \(h = a\), \(i = 3\)

Now substituting these values into the determinant formula:

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

### Step 3: Simplify the determinant
Calculating each term:

1. \(2(6 - a) = 12 - 2a\)
2. \(-a(3 - 2) = -a\)
3. \(6(a - 4) = 6a - 24\)

Combining these, we have:

\[ \text{det}(A) = (12 - 2a) - a + (6a - 24) \]

Combining like terms:

\[ \text{det}(A) = 12 - 2a - a + 6a - 24 = 3a - 12 \]

### Step 4: Set the determinant not equal to zero
For the system to have a unique solution, we require:

\[ 3a - 12 \neq 0 \]

### Step 5: Solve for 'a'
Setting the determinant equal to zero gives:

\[ 3a - 12 = 0 \implies 3a = 12 \implies a = 4 \]

Thus, for the system to have a unique solution, \(a\) cannot be equal to 4.

### Final Answer
The value of 'a' that cannot be equal to for the system to have a unique solution is:

\[ \boxed{4} \]