= Solution
The three <eigenvectors> form an <orthonormal basis>, so an arbitrary <vector> has expansion $\mathbf r=a\mathbf v_-+b\mathbf v_0+c\mathbf v_+$. Multiplication by $A$ removes the zero-eigenvalue component:
$$
A\mathbf r=-6a\mathbf v_-+12c\mathbf v_+.
$$
Its <dot product> with $\mathbf v_0$ is zero for every $a,b,c$. Thus one convenient answer is the nonzero <vector>
$$
\boxed{\mathbf u=(1,2,1)^{\mathsf T}.}
$$
Equivalently, $A^{\mathsf T}\mathbf u=A\mathbf u=0$, so $(A\mathbf r)\cdot\mathbf u=\mathbf r\cdot(A^{\mathsf T}\mathbf u)=0$. Any nonzero scalar multiple of this <vector> in the <null space> also works.
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