If $a, b, c$ are position vectors of vertices of a triangle $ABC$,then the unit vector perpendicular to its plane is:

Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b},$ where $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}.$

$i \times (j \times k) = $

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:

The unit vector perpendicular to $3i + 2j - k$ and $12i + 5j - 5k$ is

Let $\overrightarrow{OA}=\overrightarrow{a}$,$\overrightarrow{OB}=12 \overrightarrow{a}+4 \overrightarrow{b}$,and $\overrightarrow{OC}=\overrightarrow{b}$,where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$,then the ratio of the area of the quadrilateral $OABC$ to the area of $S$ is: