Given 2 vectors in 3d find a condition that satisfies an inequality.

Question

given $u=[1, 0, 1]$ and $v=[1, 2, -1]$. Let $x = sv + t\sqrt{2} \ u \times v$, where $s ∈ R$ and $t ∈ R$. Find the condition on $s$ and $t$ that renders $||x|| \leq \sqrt6$. Answer should be a relation between $s ∈ R$ and $t ∈ R$. I'm not sure if my calculation is correct. i found the cross product of $u$ and $v$ and it's norm and the norm of $v$ and substituted it into the formula and got $s^2 + t^2 \leq 1$. Is this the right approach?