Let $\vec u=u_1\hat i+u_2\hat j$ be a unit vector in xy plane and $\vec w=\frac{1}{\sqrt{6}}(\hat i+\hat j+2\hat k)$. Given that there exists a vector $\vec v$ in $R^3$ such that $|\vec u×\vec v|=1$ and $\vec w(\vec u×\vec v)=1$. Then

$|u_1|=|u_2|$

We observe that $\vec w$ is a unit vector.

Now, $|\vec u×\vec v|=1$ and $\vec w(\vec u×\vec v)=1$

$⇒\vec w\begin{vmatrix}\hat i&\hat j&\hat k\\u_1&u_2&0\\v_1&v_2&v_3\end{vmatrix}$, where $\vec v-v_1\hat i+v_2\hat j+v_3\hat k$

$⇒\frac{1}{\sqrt{6}}(\hat i+\hat j+2\hat k)=u_2\,v_3\hat i-u_1\,v_3\hat j+(u_1\,v_2-u_2\,v_1)\hat k$

$⇒u_2\,v_3=\frac{1}{\sqrt{6}},-u_1\,v_3=\frac{1}{\sqrt{6}}$ and $\frac{2}{\sqrt{6}}=u_1\,v_2-u_2\,v_1$

$⇒u_2\,v_3=-u_1\,v_3$

$⇒u_2=-u_1$

$⇒|u_2|=|-u_1|$

$⇒|u_1|=|u_2|$