Let $$f, g : \mathbb{N} - \{1\} \to \mathbb{N}$$ be functions defined by $$f(a) = \alpha$$, where $$\alpha$$ is the maximum of the powers of those primes $$p$$ such that $$p^\alpha$$ divides $$a$$, and $$g(a) = a + 1$$, for all $$a \in \mathbb{N} - \{1\}$$. Then, the function $$f + g$$ is

Given $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$ and $$\vec{b} = \hat{i} + \hat{j} - \hat{k}$$.

The first vector is $$(\vec{a} \cdot \hat{i})\hat{i} + (\vec{a} \cdot \hat{j})\hat{j} + (\vec{a} \cdot \hat{k})\hat{k}$$. Since $$\vec{a} \cdot \hat{i} = 1$$, $$\vec{a} \cdot \hat{j} = 1$$, $$\vec{a} \cdot \hat{k} = 1$$, this gives $$\hat{i} + \hat{j} + \hat{k} = \vec{a}$$.

The second vector is $$(\vec{b} \cdot \hat{i})\hat{i} + (\vec{b} \cdot \hat{j})\hat{j} + (\vec{b} \cdot \hat{k})\hat{k}$$. Since $$\vec{b} \cdot \hat{i} = 1$$, $$\vec{b} \cdot \hat{j} = 1$$, $$\vec{b} \cdot \hat{k} = -1$$, this gives $$\hat{i} + \hat{j} - \hat{k} = \vec{b}$$.

The third vector is $$\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$$.

So we have three vectors: $$\vec{a} = (1, 1, 1)$$, $$\vec{b} = (1, 1, -1)$$, $$\vec{c} = (1, 1, -2)$$.

To check coplanarity, we compute the scalar triple product $$[\vec{a}\;\vec{b}\;\vec{c}]$$:

$$[\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & 1 & -2 \end{vmatrix}$$

Expanding along the first row: $$= 1(1 \cdot (-2) - (-1) \cdot 1) - 1(1 \cdot (-2) - (-1) \cdot 1) + 1(1 \cdot 1 - 1 \cdot 1)$$

$$= 1(-2 + 1) - 1(-2 + 1) + 1(0) = (-1) - (-1) + 0 = 0$$

Since the scalar triple product is zero, the three vectors are coplanar.

The correct answer is Option B: the vectors are coplanar.