1. **State the problem:** We are given three vectors $\mathbf{u} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$, $\mathbf{v} = -2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$, and $\mathbf{w} = \mathbf{i} - 3\mathbf{j} + 5\mathbf{k}$. We need to prove that these vectors are coplanar. 2. **Formula and rule:** Three vectors are coplanar if the scalar triple product is zero. The scalar triple product is given by: $$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = 0$$ If this equals zero, the vectors lie in the same plane. 3. **Calculate the cross product $\mathbf{v} \times \mathbf{w}$:** $$\mathbf{v} = (-2, 3, -4), \quad \mathbf{w} = (1, -3, 5)$$ $$\mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix}$$ $$= \mathbf{i}(3 \times 5 - (-4) \times (-3)) - \mathbf{j}(-2 \times 5 - (-4) \times 1) + \mathbf{k}(-2 \times (-3) - 3 \times 1)$$ $$= \mathbf{i}(15 - 12) - \mathbf{j}(-10 + 4) + \mathbf{k}(6 - 3)$$ $$= 3\mathbf{i} - (-6)\mathbf{j} + 3\mathbf{k} = 3\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}$$ 4. **Calculate the dot product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$:** $$\mathbf{u} = (1, -2, 3), \quad \mathbf{v} \times \mathbf{w} = (3, 6, 3)$$ $$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = 1 \times 3 + (-2) \times 6 + 3 \times 3 = 3 - 12 + 9 = 0$$ 5. **Conclusion:** Since the scalar triple product is zero, the vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are coplanar. **Final answer:** The vectors are coplanar because $$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = 0$$.