If $A(3,2,-1), B(-2,2,-3)$ and $D(-2,5,-4)$ are the vertices of a parallelogram,then the area of the parallelogram is

Let $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k}$ and $\overrightarrow{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k}$,where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15(\alpha^{2} + 4)}$,then the value of $2|\vec{a}|^{2} + (\vec{a} \cdot \vec{b})|\vec{b}|^{2}$ is equal to

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$ and the plane determined by vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The angle between $\vec{a}$ and $(\hat{i}-2\hat{j}+2\hat{k})$ is

Let $\vec{a}=6 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $|\vec{c}| \geq 6$,$\vec{a} \cdot \vec{c}=6|\vec{c}|$,$|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^{\circ}$,then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to:

The area of the triangle having vertices as $i - 2j + 3k,$ $- 2i + 3j - k,$ and $4i - 7j + 7k$ is

If $\theta$ is the angle between the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$,find the value of $\sin \theta$.