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

Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$. Let $\vec c$ be a vector such that $|\vec c - \vec a| = 3$,$|(\vec a \times \vec b) \times \vec c| = 3$,and the angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$. Then $\vec a \cdot \vec c$ is equal to:

If $a = 2i + 4j - 5k$ and $b = i + 2j + 3k$,then $|a \times b|$ is

Let $2\hat{a} = \hat{b} \times \hat{c} + 2\hat{b}$. Then the sum of possible value$(s)$ of $\left| 2\hat{a} + \hat{b} + \hat{c} \right|$ is:

The points $A(a), B(b), C(c)$ will be collinear if

If $|a| = 4$,$|b| = 2$ and the angle between $a$ and $b$ is $\frac{\pi}{6}$,then $|a \times b|^2$ is equal to