If the area of a parallelogram,whose diagonals are $\vec{d_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{d_2} = 2\hat{i} + 3\hat{j} + \alpha\hat{k}$,is $\frac{\sqrt{93}}{2}$ sq. unit,then $\alpha = $

The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

$A$ non-zero vector $a$ is parallel to the line of intersection of the plane determined by the vectors $i, i + j$ and the plane determined by the vectors $i - j, i + k.$ The angle between $a$ and the vector $i - 2j + 2k$ is

If $a$ and $b$ are two non-zero perpendicular vectors,then a vector $y$ satisfying the equations $a \cdot y = c$ (where $c$ is a scalar) and $a \times y = b$ is

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{3}$,$|\vec{b}|=5$,$\vec{b} \cdot \vec{c}=10$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$. If $\vec{a}$ is perpendicular to the vector $\vec{b} \times \vec{c}$,then $|\vec{a} \times (\vec{b} \times \vec{c})|$ is equal to:

If $\bar{a}=\hat{i}+\hat{j}$ and $\bar{b}=2 \hat{i}-\hat{k}$,then the point of intersection of the lines $\bar{r} \times \bar{a}=\bar{b} \times \bar{a}$ and $\bar{r} \times \bar{b}=\bar{a} \times \bar{b}$ is