1. **Problem Statement:** Find the area of the parallelogram whose adjacent sides are given by vectors $\vec{a} = \hat{i} - \hat{j} + 3 \hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. 2. **Formula and Explanation:** The area of the parallelogram formed by two vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product: $$\text{Area} = |\vec{a} \times \vec{b}|$$ The cross product $\vec{a} \times \vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$, and its magnitude equals the area of the parallelogram. 3. **Calculate the cross product:** \[ \vec{a} = \begin{pmatrix}1 \\ -1 \\ 3\end{pmatrix}, \quad \vec{b} = \begin{pmatrix}2 \\ -7 \\ 1\end{pmatrix} \] \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = \hat{i} \begin{vmatrix} -1 & 3 \\ -7 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 2 & -7 \end{vmatrix} \] Calculate each minor: \[ \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2)) \] \[ = \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2) = \hat{i}(20) - \hat{j}(-5) + \hat{k}(-5) \] \[ = 20\hat{i} + 5\hat{j} - 5\hat{k} \] 4. **Find the magnitude:** $$|\vec{a} \times \vec{b}| = \sqrt{20^2 + 5^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450} = 15\sqrt{2}$$ 5. **Final answer:** The area of the parallelogram is $15\sqrt{2}$.