For the system of linear equations
$$2x + 4y + 2az = b$$
$$x + 2y + 3z = 4$$
$$2x + 5y + 2z = 8$$
which of the following is NOT correct?

Given the system:

$$2x + 4y + 2az = b \quad \cdots(1)$$

$$x + 2y + 3z = 4 \quad \cdots(2)$$

$$2x + 5y + 2z = 8 \quad \cdots(3)$$

Determinant of coefficient matrix:

$$\Delta = \begin{vmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & 5 & 2 \end{vmatrix} = 2(4-15) - 4(2-6) + 2a(5-4) = -22 + 16 + 2a = 2a - 6$$

$$\Delta = 0$$ when $$a = 3$$.

For $$a \neq 3$$: System has a unique solution regardless of $$b$$.

For $$a = 3$$: Equation (1) becomes $$2x + 4y + 6z = b$$, i.e., $$2(x + 2y + 3z) = b$$.

From (2): $$x + 2y + 3z = 4$$, so (1) requires $$b = 8$$.

Checking each option:

Option A: $$a = b = 6$$: $$\Delta = 6 \neq 0$$. Unique solution. ✓ (Correct statement)

Option B: $$a = 3, b = 6$$: $$\Delta = 0$$, but (1) gives $$8 = 6$$, contradiction. No solution, NOT infinitely many. ✗ (Incorrect statement)

Option C: $$a = 3, b = 8$$: $$\Delta = 0$$, consistent. Infinitely many solutions. ✓

Option D: $$a = b = 8$$: $$\Delta = 10 \neq 0$$. Unique solution. ✓

The statement that is NOT correct is Option B.