The ordered pair $$(a, b)$$, for which the system of linear equations
$$3x - 2y + z = b$$
$$5x - 8y + 9z = 3$$
$$2x + y + az = -1$$
has no solution, is

We need to find the ordered pair $$(a, b)$$ for which the system has no solution:

$$3x - 2y + z = b$$

$$5x - 8y + 9z = 3$$

$$2x + y + az = -1$$

Since the system has no solution, the determinant of the coefficient matrix must be zero. Thus

$$D = \begin{vmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{vmatrix}$$

$$= 3(-8a - 9) + 2(5a - 18) + 1(5 + 16)$$

$$= -24a - 27 + 10a - 36 + 21$$

$$= -14a - 42$$

Setting $$D = 0$$ gives $$-14a - 42 = 0$$, which implies $$a = -3$$.

Next, for the system to have no solution rather than infinitely many, at least one of $$D_x, D_y, D_z$$ must be non-zero. Substituting $$a = -3$$ into the determinant $$D_x$$ formed by replacing the first column of the coefficient matrix with the constants, we have

$$D_x = \begin{vmatrix} b & -2 & 1 \\ 3 & -8 & 9 \\ -1 & 1 & -3 \end{vmatrix}$$

$$= b(24 - 9) + 2(-9 + 9) + 1(3 - 8)$$

$$= 15b + 0 - 5 = 15b - 5$$

Since $$D_x \neq 0$$ is required, it follows that $$b \neq \frac{1}{3}$$.

From the given options with $$a = -3$$, namely Option B $$(-3, \frac{1}{3})$$ and Option C $$(-3, -\frac{1}{3})$$, only $$b = -\frac{1}{3}$$ satisfies this condition. Substituting $$b = -\frac{1}{3}$$ into $$D_x$$ yields

$$D_x = 15\bigl(-\tfrac{1}{3}\bigr) - 5 = -5 - 5 = -10 \neq 0$$ ✓

Therefore, the ordered pair is $$\left(-3, -\frac{1}{3}\right)$$.

The correct answer is Option C.