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

We need to find the values of $$N$$ for which the system has a unique solution, then compute $$k$$ and the sum.

The system is: $$x + y + z = 1$$, $$2x + Ny + 2z = 2$$, $$3x + 3y + Nz = 3$$.

The system has a unique solution when the determinant of the coefficient matrix is non-zero. The determinant is given by $$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N \end{vmatrix}$$.

Expanding along the first row:

$$D = 1(N^2 - 6) - 1(2N - 6) + 1(6 - 3N)$$

$$= N^2 - 6 - 2N + 6 + 6 - 3N$$

$$= N^2 - 5N + 6$$

$$= (N-2)(N-3)$$

For a unique solution, $$D \neq 0$$, so $$N \neq 2$$ and $$N \neq 3$$. Since $$N$$ can take values from 1 to 6 on a fair die, the values that give a unique solution are $$N \in \{1, 4, 5, 6\}$$, so the probability is $$\frac{4}{6}$$ and hence $$k = 4$$.

The sum of $$k$$ and all possible values of $$N$$ is calculated by adding $$4$$ and the numbers $$1, 4, 5, 6$$: $$4 + 1 + 4 + 5 + 6 = 20$$.

The answer is Option 3: 20.