Let $$N$$ denote the number that turns up when a fair die is rolled. If the probability that the system of equations
$$x + y + z = 1$$
$$2x + Ny + 2z = 2$$
$$3x + 3y + Nz = 3$$
has unique solution is $$\frac{k}{6}$$, then the sum of value of $$k$$ and all possible values of $$N$$ is

Given the system of equations where $$N$$ is the outcome of rolling a fair die ($$N \in \{1, 2, 3, 4, 5, 6\}$$):

$$x + y + z = 1 \quad \cdots(1)$$ $$2x + Ny + 2z = 2 \quad \cdots(2)$$ $$3x + 3y + Nz = 3 \quad \cdots(3)$$ $$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N \end{vmatrix}$$

Expanding along $$R_1$$:

$$\Delta = 1(N^2 - 6) - 1(2N - 6) + 1(6 - 3N)$$ $$= N^2 - 6 - 2N + 6 + 6 - 3N$$ $$= N^2 - 5N + 6$$ $$= (N - 2)(N - 3)$$

The system has a unique solution when $$\Delta \neq 0$$, i.e., $$(N-2)(N-3) \neq 0$$.

So $$N \neq 2$$ and $$N \neq 3$$.

From the die outcomes $$\{1, 2, 3, 4, 5, 6\}$$, the values giving unique solution are $$\{1, 4, 5, 6\}$$.

Number of favorable outcomes = 4.

$$P(\text{unique solution}) = \frac{4}{6} = \frac{k}{6}$$

So $$k = 4$$.

Sum of $$k$$ and all possible values of $$N$$ (that give unique solution):

$$= k + 1 + 4 + 5 + 6 = 4 + 16 = 20$$

Therefore, the answer is Option C: $$\mathbf{20}$$.