Let $\overrightarrow{c}$ be a vector perpendicular to the vectors $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}.$ If $\overrightarrow{c}\cdot(\hat{i}+\hat{j}+3\hat{k})=8,$ then the value of $\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b})$ is equal to ...... .

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $\overline{a} \neq \overline{0}$ and $\overline{a} \times \overline{b} = 2 \overline{a} \times \overline{c}$,$|\overline{a}| = |\overline{c}| = 1$,$|\overline{b}| = 4$ and $|\overline{b} \times \overline{c}| = \sqrt{15}$. If $\overline{b} - 2 \overline{c} = \lambda \overline{a}$,then $\lambda$ is

The position vectors of the points $A, B, C$ are $\hat{i}+2\hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}$,and $2\hat{i}+3\hat{j}+2\hat{k}$ respectively. If $A$ is chosen as the origin,then the cross product of the position vectors of $B$ and $C$ is:

Vectors $\vec{p}=a \hat{i}+b \hat{j}+c \hat{k}$,$\vec{q}=d \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}$ form a triangle $ABC$ such that $\vec{p}=\vec{q}+\vec{r}$. If the area of $\triangle ABC$ is $5 \sqrt{6}$ sq. units,then the sum of the absolute values of $a, b, c$ is

Let $\overrightarrow{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=((\overrightarrow{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

$a \times (b \times c)$ is coplanar with