If $\overrightarrow{\mathrm{A}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ are two vectors, then the unit vector is

$\begin{array}{l}\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 2& 1& 1\\ 1& 1& 1\end{array}\right|\\ =\widehat{\mathrm{i}}(1-1)-\widehat{\mathrm{j}}(2-1)+\widehat{\mathrm{k}}(2-1)=-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}\end{array}$

The unit vector perpendicular to $\overrightarrow{\mathrm{A}}\text{ and }\overrightarrow{\mathrm{B}}\text{ is }\left(\frac{-\widehat{\mathrm{j}}+\widehat{\mathrm{k}}}{\sqrt{2}}\right)$. So choices (a) and (c) are correct.
Any vector whose magnitude is K (constant) times $(2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}})$ is parallel to $\overrightarrow{\mathrm{A}}$.
So, unit vector $\frac{2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}}}{\sqrt{6}}$ is parallel to $\overrightarrow{\mathrm{A}}$.
So, choice (b) is correct.