$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ 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 $|\vec{a} \times(\vec{b} \times \vec{c})|=$

If $\bar{a}=2\hat{i}+3\hat{j}-\hat{k}$,$\bar{b}=-\hat{i}+2\hat{j}-4\hat{k}$ and $\bar{c}=\hat{i}+\hat{j}+\hat{k}$,then $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$

Let $\overline{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}$,$\overline{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k}$ and $\overline{c}=\hat{i}+2 \hat{j}-2 \hat{k}$,where $\alpha, \beta \in R$,be three vectors. If the projection of $\overline{a}$ on $\overline{c}$ is $\frac{10}{3}$ and $\overline{b} \times \overline{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}$,then the value of $\alpha^2+\beta^2-\alpha \beta$ is equal to

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two perpendicular lines,then the direction cosines of the line which is perpendicular to both the lines will be:

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

Let $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$,$\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c}=\vec{c} \times \vec{b}$ and $(\overrightarrow{a}+\overrightarrow{c}) \cdot(\overrightarrow{b}+\overrightarrow{c})=168$. Then the maximum value of $|\vec{c}|^2$ is :