If $a = i + j - k$,$b = i - j + k$,and $c = i - j - k$,then $a \times (b \times c) = \dots$

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $\vec{a}=\vec{b} \times(\vec{b} \times \vec{c}) .$ If magnitudes of the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are $\sqrt{2}, 1$ and $2$ respectively and the angle between $\vec{b}$ and $\vec{c}$ is $\theta$ $(0 < \theta < \frac{\pi}{2})$,then the value of $1+\tan \theta$ is equal to:

If $\overline{a}=\frac{1}{\sqrt{10}}(4 \hat{i}-3 \hat{j}+\hat{k})$ and $\overline{b}=\frac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k})$,then the value of $(2 \bar{a}-\bar{b}) \cdot \{(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b})\}$ is

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

If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a}=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}$,then the value of $(2 \vec{a}+\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})]$ is

Statement $(A)$ : If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,then $\vec{a} \times (\vec{b} \times \vec{c}) = 0$.
Reason $(R)$ : If $\vec{b}$ is perpendicular to $\vec{c}$,then $\vec{b} \times \vec{c} = 0$.