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Solution

$\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{k} + \hat{j}$. The vector perpendicular to both is $\vec{a} \times \vec{b}$. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k}$. Two unit vectors along $\pm(\hat{i} - \hat{j} + \hat{k})$ are possible.