The area of the quadrilateral $ABCD$ with vertices $A(2, 1, 1)$,$B(1, 2, 5)$,$C(-2, -3, 5)$,and $D(1, -6, -7)$ is equal to

For any two vectors $a$ and $b$,if $a \times b = 0$,then

If $\overrightarrow{OA} = 3i + 2j - k$ and $\overrightarrow{OB} = i + 3j + k$,then the area of the triangle $OAB$ is

$A (2,6,2), B (-4,0, \lambda), C (2,3,-1)$ and $D (4,5,0)$,with $|\lambda| \leq 5$,are the vertices of a quadrilateral $ABCD$. If its area is $18$ square units,then $5-6 \lambda$ is equal to $.........$.

Find the unit vector perpendicular to both vectors $\vec{a}$ and $\vec{b}$.

Let $\vec{a}=2\hat{i}-5\hat{j}+5\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+3\hat{k}$. If $\vec{c}$ is a vector such that $2(\vec{a}\times\vec{c})+3(\vec{b}\times\vec{c})=\vec{0}$ and $(\vec{a}-\vec{b})\cdot\vec{c}=-97$,then $|\vec{c}\times \hat{k}|^{2}$ is equal to