If the vectors $\vec{a} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j}$ represent the diagonals of a parallelogram,then its area will be:

If $a = i + j + 2k$ and $b = 3i + j + k$,then the value of $a \times b$ is:

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 = $

$A$ unit vector perpendicular to both the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$ is

Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$ and $\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ be two vectors such that $|\vec{a}|=1$,$\vec{a} \cdot \vec{b}=2$,and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$,then the angle between $\vec{b}$ and $\vec{c}$ is equal to:

The area of the triangle whose vertices are $(1, 2, 3)$,$(2, 5, -1)$,and $(-1, 1, 2)$ is: