The number of vectors of unit length perpendicular to vectors $\vec{a}=(1,1,0)$ and $\vec{b}=(0,1,1)$ is:

A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product $\vec{a}\times \vec{b}$.
$$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(1-0)=\hat{i}-\hat{j}+\hat{k}$$
The magnitude is $|\vec{a}\times \vec{b}|=\sqrt{1^2+(-1{)}^2+1^2}=\sqrt{3}$.Unit vectors perpendicular to both are:
$$\hat{n}=\pm \frac{\vec{a}\times \vec{b}}{|\vec{a}\times \vec{b}|}=\pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$$
There are exactly two such unit vectors (one in each direction along the normal line).