If the system of linear equations
$$x + ay + z = 3$$
$$x + 2y + 2z = 6$$
$$x + 5y + 3z = b$$
has no solution, then:

We have the system

$$x + ay + z = 3 \quad -(1)$$

$$x + 2y + 2z = 6 \quad -(2)$$

$$x + 5y + 3z = b \quad -(3)$$

For a system of three linear equations to be inconsistent (that is, to have no solution), it is enough to find two equations that contradict each other after eliminating one variable. We therefore begin by eliminating $$x$$ from the second and third equations using the first equation.

Subtract equation (1) from equation (2):

$$\bigl(x + 2y + 2z\bigr) - \bigl(x + ay + z\bigr) = 6 - 3.$$

This gives

$$\bigl(2 - a\bigr)y + \bigl(2 - 1\bigr)z = 3,$$

so

$$\boxed{(2 - a)y + z = 3}. \quad -(4)$$

Next, subtract equation (1) from equation (3):

$$\bigl(x + 5y + 3z\bigr) - \bigl(x + ay + z\bigr) = b - 3.$$

This yields

$$\bigl(5 - a\bigr)y + \bigl(3 - 1\bigr)z = b - 3,$$

so

$$\boxed{(5 - a)y + 2z = b - 3}. \quad -(5)$$

Now the original three-variable system will have no solution exactly when the two-variable subsystem (4)-(5) is itself inconsistent, because in that case there is no pair $$(y,z)$$ that can satisfy both, and consequently no value of $$x$$ can rescue the situation.

For two linear equations in $$y$$ and $$z$$,

$$a_1y + b_1z = c_1,$$

$$a_2y + b_2z = c_2,$$

there is no solution when the coefficients of $$y$$ and $$z$$ are proportional while the constants are not proportional. Symbolically, the condition for inconsistency is

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}.$$

Comparing with (4)-(5), we identify

$$a_1 = 2 - a,\quad b_1 = 1,\quad c_1 = 3,$$

$$a_2 = 5 - a,\quad b_2 = 2,\quad c_2 = b - 3.$$

First impose the proportionality of the coefficients of $$y$$ and $$z$$:

$$\frac{2 - a}{5 - a} = \frac{1}{2}.$$

Cross-multiplying, we get

$$2\,(2 - a) = 1\,(5 - a) \;\Longrightarrow\; 4 - 2a = 5 - a.$$

Bringing all terms to one side,

$$-2a + a = 5 - 4 \;\Longrightarrow\; -a = 1 \;\Longrightarrow\; a = -1.$$

So the proportionality of the coefficients forces

$$a = -1.$$

Next we require that the constants fail to have the same ratio, i.e.

$$\frac{c_1}{c_2} = \frac{3}{\,b - 3\,} \neq \frac{1}{2}.$$

Setting $$\dfrac{3}{b - 3} = \dfrac{1}{2}$$ and solving would give

$$2 \cdot 3 = 1 \cdot (b - 3) \;\Longrightarrow\; 6 = b - 3 \;\Longrightarrow\; b = 9.$$

Therefore, to ensure the inequality we must exclude this value:

$$b \neq 9.$$

Collecting the two conditions, we have

$$a = -1 \quad\text{and}\quad b \neq 9.$$

This matches Option D in the given list.

Hence, the correct answer is Option D.