$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the planes defined by the vectors $\vec{i}, \vec{i} + \vec{j}$ and $\vec{i} - \vec{j}, \vec{i} + \vec{k}$. The angle between $\vec{a}$ and the vector $\vec{i} - 2\vec{j} + 2\vec{k}$ is .....

If the diagonals of a parallelogram are represented by the vectors $3\hat{i} + \hat{j} - 2\hat{k}$ and $\hat{i} + 3\hat{j} - 4\hat{k}$,then its area in square units is:

Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}$,$\vec{b}=-\hat{i}+\hat{k}$,and $\vec{c}=\beta \hat{j}-\hat{k}$,where $\alpha$ and $\beta$ are integers and $\alpha \beta=-6$. Let the values of the ordered pair $(\alpha, \beta)$ for which the area of the parallelogram with diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$,be $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to

If the vertices of $\Delta ABC$ are $A=(2,3,5), B=(-1,3,2), C=(3,5,-2)$,then the area of the $\Delta ABC$ (in sq. units) 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:

$A$ unit vector which is coplanar to vectors $i + j + 2k$ and $i + 2j + k$ and perpendicular to $i + j + k$ is