Let the vectors $$\vec{a} = (1+t)\hat{i} + (1-t)\hat{j} + \hat{k}$$, $$\vec{b} = (1-t)\hat{i} + (1+t)\hat{j} + 2\hat{k}$$ and $$\vec{c} = t\hat{i} - t\hat{j} + \hat{k}$$, $$t \in \mathbb{R}$$ be such that for $$\alpha, \beta, \gamma \in \mathbb{R}$$, $$\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0} \Rightarrow \alpha = \beta = \gamma = 0$$. Then, the set of all values of $$t$$ is

We need to find all values of $$t$$ for which the vectors $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are linearly independent (i.e., $$\alpha\vec{a} + \beta\vec{b} + \gamma\vec{c} = \vec{0}$$ implies $$\alpha = \beta = \gamma = 0$$).

Since the vectors are linearly independent if and only if the determinant of the matrix formed by their components is non-zero, we consider:

$$D = \begin{vmatrix} 1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ t & -t & 1 \end{vmatrix}$$

Expanding this determinant along the first row gives:

$$D = (1+t)[(1+t)(1) - 2(-t)] - (1-t)[(1-t)(1) - 2t] + 1[(-t)(1-t) - t(1+t)]$$

This simplifies to:

$$= (1+t)(1+t+2t) - (1-t)(1-t-2t) + (-t+t^2 - t - t^2)$$

$$= (1+t)(1+3t) - (1-t)(1-3t) + (-2t)$$

$$= (1 + 3t + t + 3t^2) - (1 - 3t - t + 3t^2) - 2t$$

$$= (1 + 4t + 3t^2) - (1 - 4t + 3t^2) - 2t$$

$$= 8t - 2t = 6t$$

From the above, the determinant is $$D = 6t$$, so the vectors are linearly independent exactly when $$D \neq 0$$, that is, when $$6t \neq 0$$, or equivalently $$t \neq 0$$.

Therefore, the set of all such values of $$t$$ is $$\mathbb{R} - \{0\}$$.

The answer is Option C.