1. **State the problem:** We need to find a unit vector $\mathbf{u}$ orthogonal to vectors $\mathbf{a} = \mathbf{i} + \mathbf{j} - \mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} - \mathbf{j}$. The vector $\mathbf{u}$ must satisfy $\mathbf{u} \cdot \mathbf{a} = 0$ and $\mathbf{u} \cdot \mathbf{b} = 0$, and have magnitude 1. 2. **Find the vector orthogonal to both $\mathbf{a}$ and $\mathbf{b}$:** The cross product $\mathbf{a} \times \mathbf{b}$ gives a vector orthogonal to both. $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 2 & -1 & 0 \end{vmatrix} = \mathbf{i}(1 \cdot 0 - (-1)(-1)) - \mathbf{j}(1 \cdot 0 - (-1)(2)) + \mathbf{k}(1 \cdot (-1) - 1 \cdot 2)$$ $$= \mathbf{i}(0 - 1) - \mathbf{j}(0 + 2) + \mathbf{k}(-1 - 2) = -\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}$$ 3. **Calculate the magnitude of $\mathbf{a} \times \mathbf{b}$:** $$|\mathbf{a} \times \mathbf{b}| = \sqrt{(-1)^2 + (-2)^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$ 4. **Find the unit vector $\mathbf{u}$:** $$\mathbf{u} = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} = \left(-\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, -\frac{3}{\sqrt{14}}\right)$$ 5. **Compare with given options:** Option (a) is $\left(-\frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, -\frac{3}{\sqrt{6}}\right)$ which is not equal to our result. Option (b) is $(-1, 2, -3)$ which is not a unit vector and also differs in sign for the $j$ component. Therefore, **(c) None of the above is true** is correct. **Final answer:** (c) None of the above is true.