Let $$\vec{a} = \hat{i} + \hat{j} - \hat{k}$$ and $$\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}$$. Then the number of vectors $$\vec{b}$$ such that $$\vec{b} \times \vec{c} = \vec{a}$$ and $$|\vec{b}| \in \{1, 2, \ldots, 10\}$$ is

We are given $$\vec{a} = \hat{i} + \hat{j} - \hat{k}$$ and $$\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}$$. We need to find the number of vectors $$\vec{b}$$ such that $$\vec{b} \times \vec{c} = \vec{a}$$ and $$|\vec{b}| \in \{1, 2, \ldots, 10\}$$.

First, for the cross product equation $$\vec{b} \times \vec{c} = \vec{a}$$ to have a solution, $$\vec{a}$$ must be perpendicular to $$\vec{c}$$ because $$\vec{b} \times \vec{c}$$ is always perpendicular to $$\vec{c}$$.

Now, computing the dot product gives $$\vec{a} \cdot \vec{c} = (1)(2) + (1)(-3) + (-1)(2) = 2 - 3 - 2 = -3 \neq 0$$.

Since $$\vec{a} \cdot \vec{c} \neq 0$$, $$\vec{a}$$ is not perpendicular to $$\vec{c}$$, and consequently there is no vector $$\vec{b}$$ satisfying $$\vec{b} \times \vec{c} = \vec{a}$$.

The number of such vectors $$\vec{b}$$ is $$0$$.

The answer is Option A: $$0$$.