The system of linear equations
$$x + y + z = 6$$
$$2x + 5y + az =36$$
$$x + 2y + 3z = b$$

We are given the system of linear equations:

$$x + y + z = 6 \quad \cdots (1)$$

$$2x + 5y + az = 36 \quad \cdots (2)$$

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

The coefficient matrix is:

$$A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix}$$

Its determinant is

$$\det(A) = 1(15 - 2a) - 1(6 - a) + 1(4 - 5) = 15 - 2a - 6 + a - 1 = 8 - a.$$

Since $$\det(A) = 8 - a\,, $$ for $$a = 8$$ we have $$\det(A) = 0$$, so the system does not have a unique solution and Options 1 and 3 are eliminated.

Substituting $$a = 8$$ into the augmented matrix gives

$$\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 5 & 8 & | & 36 \\ 1 & 2 & 3 & | & b \end{pmatrix}.$$

Applying $$R_2 \to R_2 - 2R_1$$ yields $$(0, 3, 6 \mid 24)$$ and then $$R_3 \to R_3 - R_1$$ yields $$(0, 1, 2 \mid b - 6).$$ Finally, $$R_3 \to R_3 - \tfrac{1}{3}R_2$$ gives $$(0, 0, 0 \mid b - 14).$$

For consistency (infinitely many solutions) we require $$b - 14 = 0\,, $$ so $$b = 14.$$ If instead $$b = 16\,, $$ then $$b - 14 = 2 \neq 0$$ and the system is inconsistent (no solution). Therefore the correct answer is Option 4: infinitely many solutions for $$a = 8$$ and $$b = 14$$.