Let $$\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$$, and $$\vec{b}$$ and $$\vec{c}$$ be two nonzero vectors such that $$\vec{a} + \vec{b} + \vec{c} = \vec{a} + \vec{b} - \vec{c}$$ and $$\vec{b} \cdot \vec{c} = 0$$. Consider the following two statements:
A: $$\vec{a} + \lambda\vec{c} \ge \vec{a}$$ for all $$\lambda \in \mathbb{R}$$.
B: $$\vec{a}$$ and $$\vec{c}$$ are always parallel.

Given $$\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$$, and the condition $$|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$$ with $$\vec{b} \cdot \vec{c} = 0$$.

From $$|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$$, squaring both sides:

$$|\vec{a} + \vec{b}|^2 + 2(\vec{a} + \vec{b}) \cdot \vec{c} + |\vec{c}|^2 = |\vec{a} + \vec{b}|^2 - 2(\vec{a} + \vec{b}) \cdot \vec{c} + |\vec{c}|^2$$

This gives $$4(\vec{a} + \vec{b}) \cdot \vec{c} = 0$$, so $$(\vec{a} + \vec{b}) \cdot \vec{c} = 0$$.

Since $$\vec{b} \cdot \vec{c} = 0$$, we get $$\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0$$, which gives $$\vec{a} \cdot \vec{c} = 0$$.

Statement A: $$|\vec{a} + \lambda\vec{c}| \geq |\vec{a}|$$ for all $$\lambda \in \mathbb{R}$$.

$$|\vec{a} + \lambda\vec{c}|^2 = |\vec{a}|^2 + 2\lambda(\vec{a} \cdot \vec{c}) + \lambda^2|\vec{c}|^2 = |\vec{a}|^2 + \lambda^2|\vec{c}|^2$$

Since $$\vec{a} \cdot \vec{c} = 0$$ and $$\lambda^2|\vec{c}|^2 \geq 0$$, we have $$|\vec{a} + \lambda\vec{c}|^2 \geq |\vec{a}|^2$$.

Therefore $$|\vec{a} + \lambda\vec{c}| \geq |\vec{a}|$$. Statement A is correct.

Statement B: $$\vec{a}$$ and $$\vec{c}$$ are always parallel.

Since $$\vec{a} \cdot \vec{c} = 0$$ and $$\vec{c} \neq \vec{0}$$, $$\vec{a}$$ and $$\vec{c}$$ are perpendicular, not parallel. Statement B is false.

The answer is Option C: only (A) is correct.