Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}-7\hat{j}+\hat{k}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} \times \vec{b} = 4\vec{c}$,$\vec{b} \times \vec{c} = 9\vec{a}$,and $\vec{c} \times \vec{a} = \alpha\vec{b}$,where $\alpha > 0$. If $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$,then $\alpha$ is equal to:

If $a+2b+3c=0$ and $(a \times b)+(b \times c)+(c \times a)=\lambda(b \times c)$,then the value of $\lambda$ is equal to

Find the unit vector perpendicular to the plane containing the points $(1, -1, 2), (2, 0, -1),$ and $(0, 2, 1)$.

For any three vectors $\vec{a}, \vec{b}, \vec{c}$,the value of $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b})$ is equal to:

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