$A$ unit vector coplanar with $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ and perpendicular to $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ is

Given $\bar{a} = 2\bar{i} + \bar{j} - 2\bar{k}$ and $\bar{b} = \bar{i} + \bar{j}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c} = |\bar{c}|$,$|\bar{c} - \bar{a}| = 2\sqrt{2}$,and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $30^{\circ}$,then the value of $|(\bar{a} \times \bar{b}) \times \bar{c}|^2$ is

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=1$,$|\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$,then $192 \sin^2 \alpha$ is equal to

$\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) = $

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\overrightarrow{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$,$\overrightarrow{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{d}$ be a vector such that $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \cdot \overrightarrow{d} = 4$. Then $|(\overrightarrow{a} \times \overrightarrow{d})|^2$ is equal to . . . . . . .

Let $m$ be a vector of magnitude $\sqrt{3}$ and perpendicular to the vectors $\hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. Let $n$ be another vector of magnitude $2\sqrt{6}$ and perpendicular to the vectors $2\hat{i}-\hat{j}$ and $\hat{j}+2\hat{k}$. The area (in sq. units) of the triangle formed with $m$ and $n$ as sides is