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

Let $2\hat{a} = \hat{b} \times \hat{c} + 2\hat{b}$. Then the sum of possible value$(s)$ of $\left| 2\hat{a} + \hat{b} + \hat{c} \right|$ is:

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar unit vectors such that $\bar{a} \times(\bar{b} \times \bar{c})=\frac{(\bar{b}+\bar{c})}{\sqrt{2}}$,then the angle between $\bar{a}$ and $\bar{b}$ is

The unit vectors perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ are:

Let $\overrightarrow{a}=\hat{i}+5\hat{j}+\alpha\hat{k}$,$\overrightarrow{b}=\hat{i}+3\hat{j}+\beta\hat{k}$ and $\overrightarrow{c}=-\hat{i}+2\hat{j}-3\hat{k}$ be three vectors such that $|\overrightarrow{b} \times \overrightarrow{c}|=5\sqrt{3}$ and $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$. Then the greatest value of $|\vec{a}|^{2}$ is .... .

If $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$,then the value of $(2\vec{a} - \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} \times 2\vec{b})]$ is: