1. Interpret the two given conditions

Let

Then the conditions become

and

Also,

Its magnitude is

so $\widehat w$ is also a unit vector.

Now for any vectors $\mathbf a,\widehat w$ with both magnitudes $1$,

Given $\widehat w\cdot \mathbf a=1$, we must have

so

Hence the problem is equivalent to asking when there exists $\overrightarrow v$ such that

2. Necessary and sufficient condition for solvability

Since $\widehat u\times \overrightarrow v$ is always perpendicular to $\widehat u$, we must have

This is necessary.

It is also sufficient: if $\widehat u\cdot \widehat w=0$, then choosing

gives [ \widehat u\times \overrightarrow v =\widehat u\times(-\widehat u\times \widehat w). ] Using

we get

Since $\widehat u\cdot \widehat w=0$ and $|\widehat u|=1$,

Therefore

So the condition is exactly

Now compute:

Thus existence of such $\overrightarrow v$ is equivalent to

3. Check options A and B

We need solutions of

Suppose $\overrightarrow v_0$ is one solution. Then for any real $\lambda$,

So infinitely many solutions exist.

Also, there cannot be exactly one solution, because adding any multiple of $\widehat u$ gives another.

Therefore:

• A is false
• B is true

4. Check option C: $\widehat u$ lies in the $xy$-plane

If $\widehat u$ lies in the $xy$-plane, then

The existence condition becomes

Hence

So C is true.

5. Check option D: $\widehat u$ lies in the $xz$-plane

If $\widehat u$ lies in the $xz$-plane, then

The existence condition becomes

So

which is equivalent to

not

Hence D is false.

6. Final answer

Correct statements are

This matches the stored correct answer.