Let $a=2 \hat{i}-2 \hat{j}+\hat{k}$ and $b=-\hat{j}+\hat{k}$. If $c$ is a vector such that $a \cdot c=|c|$,$|c-a|=2 \sqrt{2}$,and the angle between $a \times b$ and $c$ is $\frac{\pi}{3}$,then $|(a \times b) \times c|=$

$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors and $\vec{a}$ is a unit vector such that $\cos (\vec{a}, \vec{b} \times \vec{c})=\sqrt{\frac{2}{3}}$. Then $|\vec{a} \times(\vec{b} \times \vec{c})|=$

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:

$A (2,6,2), B (-4,0, \lambda), C (2,3,-1)$ 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 $.........$.

If $\overrightarrow{OA}=2 \hat{i}-\hat{j}+\hat{k}$,$\overrightarrow{OB}=3 \hat{i}-\hat{k}$ and $\overrightarrow{OC}=2 \hat{j}+3 \hat{k}$ are the position vectors of the points $A, B$ and $C$,then a unit vector perpendicular to the plane containing $A, B$ and $C$ is

$\text{If } \vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \text{ and } \vec{c} = \hat{i} - \hat{j} \text{ and if } 6\hat{i} + 2\hat{j} + 3\hat{k} = \lambda_1(\vec{a} \times \vec{b}) + \lambda_2(\vec{b} \times \vec{c}) + \lambda_3(\vec{c} \times \vec{a}), \text{ then } (\lambda_1, \lambda_2, \lambda_3) = $