linearly dependent in Q, linearly independent in R

I need help with this one. :(

Find $\displaystyle \lambda \in \mathbb{R}$ such that $\displaystyle \{(\lambda,1,0),(1,\lambda,1),(0,1,\lambda)\}

$ is linearly dependent in $\displaystyle \mathbb{Q}$ and linearly independent in $\displaystyle \mathbb{R}$.

Here's what I did:

$\displaystyle \alpha (\lambda, 1, 0)+\beta(1, \lambda, 1)+\gamma (0, 1, \lambda)=0$

Solving this gets me:

$\displaystyle \alpha \lambda+\beta=0$

$\displaystyle \alpha + \beta \lambda + \gamma=0$ and

$\displaystyle \beta + \gamma \lambda=0$.

This gets me:

$\displaystyle \lambda(\gamma - \alpha)=0$ which means that either $\displaystyle \lambda=0$ or $\displaystyle \gamma - \alpha=0$.

The first case gives me $\displaystyle \alpha, \beta, \gamma=0$, which means that the set is linearly independent both in $\displaystyle \mathbb{R}$ and $\displaystyle \mathbb{Q}$ so that is obviously not the $\displaystyle \lambda$ I'm looking for.

Now, the other case, $\displaystyle \gamma-\alpha=0$. This means that $\displaystyle \alpha=\gamma$. The three equations I had are now two equations:

$\displaystyle \alpha \lambda + \beta=0$ and

$\displaystyle 2\alpha + \beta \lambda=0$.

Solving that gets me:

$\displaystyle \alpha (2-\lambda^2)=0$.

Now, if $\displaystyle \lambda=\pm \sqrt{2} \in \mathbb{R}$, I get that $\displaystyle \alpha, \beta, \gamma $ are not necessarily zero, but if $\displaystyle \lambda \in \mathbb{Q}$, the equation $\displaystyle \alpha (2-\lambda^2)=0$ implies that $\displaystyle \alpha=0$, and so $\displaystyle \beta=0$ and $\displaystyle \gamma=0$.

Obviously, $\displaystyle \lambda=\pm \sqrt{2}$ is the one we're looking for.

Is this correct?

Is this the only possible $\displaystyle \lambda$?

Thank you!