= Solution
A <vector> perpendicular to $\mathbf e_1$ has the form $\mathbf v=(s,t,-s)$. Direct <matrix multiplication> gives
$$
A\mathbf v=(-2s,-2t,2s)=-2\mathbf v.
$$
Thus every \b[nonzero] such <vector> is an <eigenvector> with $\boxed{\lambda=-2}$. The zero <vector> must be excluded from the literal claim about “any” perpendicular <vector>, since an <eigenvector> is nonzero. The <eigenspace> is two-dimensional. Equivalently, $A=-2I+6\mathbf e_1\mathbf e_1^{\mathsf T}$ exhibits the <spectrum of a scalar matrix plus a rank-one projector>.
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