$\bar{a}=\hat{i}-\hat{\mathrm{j}}, \overline{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}, \overline{\mathrm{c}}=\hat{\mathrm{k}}-\hat{i}$ then a unit vector $\overline{\mathrm{d}}$ such that $\bar{a} \cdot \bar{d}=0=[\bar{b} \bar{c} \bar{d}]$ is

Let $\bar{d}=x\hat{i}+y\hat{j}+z\hat{k}$ be a unit vector orthogonal to $\bar{a}=\hat{i}-\hat{j}$ and coplanar with $\bar{b}=\hat{j}-\hat{k}$, $\bar{c}=\hat{k}-\hat{i}$.

The orthogonality condition $\bar{a}\cdot\bar{d}=0$ gives $x-y=0$, so $x=y$.

The coplanarity of $\bar{b}$, $\bar{c}$, $\bar{d}$ is equivalent to $\left[\bar{b}\ \bar{c}\ \bar{d}\right]=0$, which implies $\bar{d}$ is orthogonal to $\bar{b}\times\bar{c}$.

Compute $\bar{b}\times\bar{c}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ -1 & 0 & 1\end{vmatrix}=\hat{i}(1\cdot1-(-1)\cdot0)-\hat{j}(0\cdot1-(-1)\cdot(-1))+\hat{k}(0\cdot0-1\cdot(-1))=\hat{i}+\hat{j}+\hat{k}$.

Then $\bar{d}\cdot(\bar{b}\times\bar{c})=x+y+z=0$. With $x=y$, we get $2x+z=0$, so $z=-2x$.

Thus $\bar{d}=x(\hat{i}+\hat{j}-2\hat{k})$. The unit vector condition requires $|x|\sqrt{1^2+1^2+(-2)^2}=1$, so $|x|\sqrt{6}=1$ and $x=\pm\frac{1}{\sqrt{6}}$.

Therefore $\bar{d}=\pm\frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}$, which corresponds to Option D.