We have $$\widehat u = {u_1}\widehat i + {u_2}\widehat j + {u_3}\widehat k$$

That is, $$\left| {\widehat u} \right| = 1 = \sqrt {u_1^2 + u_2^2 + u_3^2} $$

$$ \Rightarrow u_1^2 + u_2^2 + u_3^2 = 1$$

Also, it is given that

$$\widehat \omega = {1 \over {\sqrt 6 }}(\widehat i + \widehat j + 2\widehat k)$$

That is, $$\left| {\widehat \omega } \right| = 1$$

Now, $$\left| {\widehat u \times \overrightarrow v } \right| = 1$$

That is, $$\left| {\widehat u} \right|\left| {\overrightarrow v } \right|\sin \theta = 1 \Rightarrow \left| {\overrightarrow v } \right| = {1 \over {\sin \theta }}$$

which shows that there are infinitely many possible values exist for $$\overrightarrow v $$ (here $$\theta$$ is angle between the vectors $$\overrightarrow v $$ and $$\widehat u$$).

Hence, option (B) is correct.

Now, $$\widehat \omega \,.\,(\widehat u \times \overrightarrow v ) = 1$$

$$\left| {\widehat \omega \,.\,(\widehat u \times \overrightarrow v )} \right| = 1$$

That is, $$\left| {\widehat \omega } \right|\left| {\widehat u \times \overrightarrow v } \right|\cos \alpha = 1$$

where $$\alpha$$ is the angle between $$\widehat \omega $$ and $$\widehat u \times \overrightarrow v $$.

Therefore, (1)(1)cos$$\alpha$$ = 1

$$\Rightarrow$$ a = 0

which means that $$\widehat \omega $$ and $$\widehat u \times \overrightarrow v $$ are parallel vector or $$\widehat \omega $$ is perpendicular vector to $$\widehat u$$ and $$\overrightarrow v $$.

$$\widehat u\,.\,\widehat \omega = 0$$

$$({u_1}\widehat i + {u_2}\widehat j + {u_3}\widehat k)\,.\,\left( {{{(\widehat i + \widehat j + 2\widehat k)} \over {\sqrt 6 }}} \right) = 0$$

$${u_1} + {u_2} + 2{u_3} = 0$$

If $$\widehat u$$ lies in xy-plane then u₃ = 0. Therefore,

$${u_1} + {u_2} = 0 \Rightarrow {u_1} = - {u_2} \Rightarrow \left| {{u_1}} \right| = \left| {{u_2}} \right|$$

Hence, option (C) is correct.