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

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

We observe that $$\vec{v} \cdot (\vec{v} \times \vec{w}) = 0$$, which implies:

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

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

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

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

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

Substituting this value of $$\lambda$$ into the given expression for $$\vec{v} \times \vec{w}$$ yields:

$$\vec{v} \times \vec{w} = \vec{u} - \frac{1}{2}\vec{v}$$

Applying the vector triple product identity gives:

$$\vec{v} \times (\vec{v} \times \vec{w}) = \vec{v}(\vec{v} \cdot \vec{w}) - \vec{w}(\vec{v} \cdot \vec{v})$$

$$= 2\vec{v} - 6\vec{w}$$

On the other hand, using the expression for $$\vec{v} \times \vec{w}$$ leads to:

$$\vec{v} \times (\vec{v} \times \vec{w}) = \vec{v} \times (\vec{u} - \frac{1}{2}\vec{v}) = \vec{v} \times \vec{u}$$

$$\vec{v} \times \vec{u} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & -1 & -2 \end{vmatrix} = \hat{i}(-2 - 1) - \hat{j}(-4 + 1) + \hat{k}(-2 - 1) = -3\hat{i} + 3\hat{j} - 3\hat{k}$$

Equating the two expressions for $$\vec{v} \times (\vec{v} \times \vec{w})$$ gives:

$$-3\hat{i} + 3\hat{j} - 3\hat{k} = 2\vec{v} - 6\vec{w} = (4\hat{i} + 2\hat{j} - 2\hat{k}) - 6\vec{w}$$

$$6\vec{w} = 4\hat{i} + 2\hat{j} - 2\hat{k} + 3\hat{i} - 3\hat{j} + 3\hat{k} = 7\hat{i} - \hat{j} + \hat{k}$$

$$\vec{w} = \frac{1}{6}(7\hat{i} - \hat{j} + \hat{k})$$

Finally, we compute:

$$\vec{u} \cdot \vec{w} = \frac{1}{6}(7(1) + (-1)(-1) + (1)(-2)) = \frac{1}{6}(7 + 1 - 2) = \frac{6}{6} = 1$$

Therefore, $$\vec{u} \cdot \vec{w} = 1$$, which corresponds to Option 1.