1. **Problem Statement:** Find a unit vector perpendicular to the plane formed by vectors $\vec{A} = 2\mathbf{i} - 3\mathbf{j} - \mathbf{k}$ and $\vec{B} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$. 2. **Formula and Concept:** A vector perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ is given by their cross product: $$\vec{N} = \vec{A} \times \vec{B}$$ The unit vector $\hat{n}$ perpendicular to the plane is: $$\hat{n} = \frac{\vec{N}}{|\vec{N}|}$$ 3. **Calculate the cross product $\vec{A} \times \vec{B}$:** $$\vec{A} = (2, -3, -1), \quad \vec{B} = (1, 4, -2)$$ $$\vec{A} \times \vec{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2 \end{vmatrix}$$ Calculate determinant: $$= \mathbf{i}((-3)(-2) - (-1)(4)) - \mathbf{j}(2(-2) - (-1)(1)) + \mathbf{k}(2(4) - (-3)(1))$$ $$= \mathbf{i}(6 + 4) - \mathbf{j}(-4 + 1) + \mathbf{k}(8 + 3)$$ $$= 10\mathbf{i} - (-3)\mathbf{j} + 11\mathbf{k} = 10\mathbf{i} + 3\mathbf{j} + 11\mathbf{k}$$ 4. **Calculate the magnitude of $\vec{N}$:** $$|\vec{N}| = \sqrt{10^2 + 3^2 + 11^2} = \sqrt{100 + 9 + 121} = \sqrt{230}$$ 5. **Find the unit vector:** $$\hat{n} = \frac{1}{\sqrt{230}} (10\mathbf{i} + 3\mathbf{j} + 11\mathbf{k})$$ 6. **Answer:** The unit vector perpendicular to the plane is $$\boxed{\frac{1}{\sqrt{230}} (10\mathbf{i} + 3\mathbf{j} + 11\mathbf{k})}$$ which corresponds to option A.