In the first case, suppose $|b| < |a|.$ Then $ | 2 a b| = \xi^2 \leq 2 a^2,$$ | 2 a b| = \xi^2 < 2 a^2,$ so in fact
$$ \left( \frac{\xi}{a} \right)^2 < 2.$$
We choose the root $ \tilde{\lambda} = \left( 0, 1, 1 \right).$ Then $z \cdot \tilde{\lambda} = b + a,$ and
$$ s_{\tilde{\lambda}} (z) = z - ( \tilde{\lambda} \cdot z) \tilde{\lambda} = (\xi, -b,-a).$$