1. Given vectors

$\vec a = \hat i + \hat j - \hat k = (1,1,-1), \qquad \vec c = 2\hat i - 3\hat j + 2\hat k = (2,-3,2)$

We need vectors $\vec b$ such that

$\vec b \times \vec c = \vec a$

and

$|\vec b| \in \{1,2,\dots,10\}.$

2. Necessary condition for a cross product equation

For any vectors $\vec b$ and $\vec c$, the vector $\vec b \times \vec c$ is always perpendicular to $\vec c$.

So if

$\vec b \times \vec c = \vec a,$

then necessarily

$\vec a \cdot \vec c = 0.$

Let us check this:

$\vec a \cdot \vec c = (1)(2) + (1)(-3) + (-1)(2) = 2 - 3 - 2 = -3.$

Since

$\vec a \cdot \vec c \neq 0,$

$\vec a$ is not perpendicular to $\vec c$.

Hence it is impossible for $\vec a$ to be equal to $\vec b \times \vec c$ for any vector $\vec b$.

3. Conclusion about existence of $\vec b$

There is no vector $\vec b$ satisfying

$\vec b \times \vec c = \vec a.$

Therefore, certainly there are no such vectors with magnitude in $\{1,2,\dots,10\}$.

So the required number of vectors is

$0.$

4. Option check

• A: 0 ✅
• B: 1 ❌
• C: 2 ❌
• D: 3 ❌

Thus the correct option is A.