To find $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}$, where $\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}$, $\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}$, and $\mathbf{c} = \mathbf{i} + \mathbf{j} - \mathbf{k}$, we will proceed step-by-step:

Step 1: Compute $\mathbf{a} \times \mathbf{b}$

Given: $\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}$ $\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}$

The cross product $\mathbf{a} \times \mathbf{b}$ is determined by the following determinant:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$ Expand this determinant: $$\mathbf{a} \times \mathbf{b} = \mathbf{i} \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}$$ Calculating each 2x2 determinant: $$\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2$$ $$\begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1) = 1 - 1 = 0$$ $$\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = (1)(-1) - (1)(1) = -1 - 1 = -2$$ Thus, the cross product is: $$\mathbf{a} \times \mathbf{b} = 2\mathbf{i} - 0\mathbf{j} - 2\mathbf{k} = 2\mathbf{i} - 2\mathbf{k}$$ ### Step 2: Compute $$(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}$$ Given: $$\mathbf{c} = \mathbf{i} + \mathbf{j} - \mathbf{k}$$ Now, calculate the dot product: $$(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (2\mathbf{i} - 2\mathbf{k}) \cdot (\mathbf{i} + \mathbf{j} - \mathbf{k})$$ Compute the dot product by multiplying corresponding components: $$(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (2)(1) + (0)(1) + (-2)(-1) = 2 + 0 + 2 = 4$$ So, the correct answer is: $$\boxed{4}$$ However, none of the provided options matches this answer, so it appears the choices may have a typo. Given our solution steps, the correct value for $$(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}$$ is indeed $$4$$.