Let $$\vec{u} = \hat{i} - \hat{j} - 2\hat{k}$$, $$\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$$, $$\vec{v} \cdot \vec{w} = 2$$ and $$\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$$, then $$\vec{u} \cdot \vec{w}$$ is equal to

$$\vec{u} = \hat{i} - \hat{j} - 2\hat{k}$$, $$\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$$, $$\vec{v} \cdot \vec{w} = 2$$, $$\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$$.

$$\vec{v} \cdot (\vec{u} + \lambda\vec{v}) = 0$$

$$\vec{v} \cdot \vec{u} + \lambda|\vec{v}|^2 = 0$$

$$\vec{v} \cdot \vec{u} = 2(1) + 1(-1) + (-1)(-2) = 3$$

$$|\vec{v}|^2 = 4 + 1 + 1 = 6$$

$$3 + 6\lambda = 0 \implies \lambda = -\frac{1}{2}$$

$$\vec{w} \cdot (\vec{u} + \lambda\vec{v}) = 0$$

$$\vec{u} \cdot \vec{w} + \lambda(\vec{v} \cdot \vec{w}) = 0$$

$$\vec{u} \cdot \vec{w} + (-\frac{1}{2})(2) = 0$$

$$\vec{u} \cdot \vec{w} = 1$$

The correct answer is Option 1: $$1$$.