Let $$\vec{a} = -\hat{i} - \hat{j} + \hat{k}$$, $$\vec{a} \cdot \vec{b} = 1$$ and $$\vec{a} \times \vec{b} = \hat{i} - \hat{j}$$. Then $$\vec{a} - 6\vec{b}$$ is equal to

Given $$\vec{a} = -\hat{i} - \hat{j} + \hat{k}$$, $$\vec{a} \cdot \vec{b} = 1$$, and $$\vec{a} \times \vec{b} = \hat{i} - \hat{j}$$.

Let $$\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$$

From $$\vec{a} \cdot \vec{b} = 1$$:

$$ -x - y + z = 1 \quad \cdots (1) $$

From $$\vec{a} \times \vec{b} = \hat{i} - \hat{j}$$:

$$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 1 \\ x & y & z \end{vmatrix} $$

$$ = \hat{i}(-z - y) - \hat{j}(-z - x) + \hat{k}(-y + x) $$

$$ = (-z-y)\hat{i} + (z+x)\hat{j} + (x-y)\hat{k} $$

Comparing with $$\hat{i} - \hat{j} + 0\hat{k}$$:

$$ -z - y = 1 \quad \cdots (2) $$

$$ z + x = -1 \quad \cdots (3) $$

$$ x - y = 0 \quad \cdots (4) $$

From (4): $$x = y$$

From (3): $$z = -1 - x$$

From (2): $$-(-1-x) - x = 1 \implies 1 + x - x = 1 \implies 1 = 1$$ ✓ (consistent)

From (1): $$-x - x + (-1-x) = 1 \implies -3x - 1 = 1 \implies x = -\frac{2}{3}$$

So $$y = -\frac{2}{3}$$, $$z = -1 + \frac{2}{3} = -\frac{1}{3}$$.

$$\vec{b} = -\frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k}$$

Compute $$\vec{a} - 6\vec{b}$$

$$ \vec{a} - 6\vec{b} = (-1 + 4)\hat{i} + (-1 + 4)\hat{j} + (1 + 2)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} $$

$$ = 3(\hat{i} + \hat{j} + \hat{k}) $$

This matches Option 2: $$3(\hat{i} + \hat{j} + \hat{k})$$.

The answer key says 25, which doesn't correspond to a standard option number. Our answer is Option 2.

The answer is $$\boxed{3(\hat{i} + \hat{j} + \hat{k})}$$.