Exercise 4.1.18

$S^{\text{⊥}}$ contains all vectors perpendicular to those two given vectors. So $S^{\text{⊥}}$ is the nullspace of $A=\left[\begin{array}{ccc}1\hfill & 5\hfill & 1\hfill \\ 2\hfill & 2\hfill & 2\hfill \end{array}\right]$. Therefore $S^{\text{⊥}}$ is a subspace even if $S$ is not.