$(b \times c) \times (c \times a) = \dots$

Let $\overrightarrow{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$,$\overrightarrow{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$,and a vector $\vec{c}$ be such that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$. If $\vec{a} \cdot \vec{c}=13$,then $(24-\vec{b} \cdot \vec{c})$ is equal to ...........

Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k}$,$\vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{i}-\hat{j}$. Then $\vec{a}-6 \vec{b}$ is equal to

$A$ unit vector perpendicular to vector $c$ and coplanar with vectors $a$ and $b$ is

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors of magnitudes $\sqrt{3}, 1, 2$ respectively,such that $\vec{a} \times (\vec{a} \times \vec{c}) + 3\vec{b} = \vec{0}$. If $\theta$ is the angle between $\vec{a}$ and $\vec{c}$,then $\cos^2 \theta = $

If $a=(1,2,3), b=(2,-1,1), c=(3,2,1)$ and $a \times(b \times c)=\alpha a+\beta b+\gamma c$,then