Find the area of the triangle with vertices $A(1,1,2), B(2,3,5)$ and $C(1,5,5)$.

Let $\bar{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$,$\bar{b} = 3 \hat{i} - \hat{j} + \beta \hat{k}$,and $\bar{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ where $\alpha, \beta \in R$,be three vectors. If the projection of $\bar{a}$ on $\bar{c}$ is $\frac{10}{3}$ and $\bar{b} \times \bar{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$,then the value of $(\alpha + \beta)$ is equal to

If $a = 2i - 3j - k$ and $b = i + 4j - 2k$,then $a \times b$ is

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

Let $\bar{a}, \bar{b}, \bar{c}$ be vectors such that $\bar{a} \neq \bar{o}, \bar{b} \neq \bar{o}, \bar{a} \times \bar{c} = \bar{b}$ and $\bar{b} \times \bar{c} = \bar{a}$. Then:

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 ...... .