If $A(3, 1, -1)$,$B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right)$,$C(2, 2, 1)$ and $D\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)$ are the vertices of a quadrilateral $ABCD$,then its area is

Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$ is equal to

If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ represent the adjacent sides of a parallelogram,then the area of this parallelogram is:

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|\vec{a} \cdot \vec{c}|$ is:

Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=\hat{i}+\hat{j}$ and $\vec{c}$ be a vector such that $|\vec{c}-\vec{a}|=3$. If $\vec{p}=\vec{a} \times \vec{b}$,then the angle between $\vec{p}$ and $\vec{c}$ is $\frac{\pi}{6}$ and $|\vec{p} \times \vec{c}|=3$. Thus,$\vec{a} \cdot \vec{c}$ is equal to:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k},$ find a vector $\vec{c}$ such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3.$