If $r=b+ta$ and $r=d+sc$ are two skew lines, then the shortest distance between them is

Let $ABC$ be a triangle such that $\overrightarrow{BC} = \overrightarrow{a}$,$\overrightarrow{CA} = \overrightarrow{b}$,$\overrightarrow{AB} = \overrightarrow{c}$,$|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$. Consider the statements:
$(S1): |(\overrightarrow{a} \times \overrightarrow{b}) + (\overrightarrow{c} \times \overrightarrow{b})| - |\overrightarrow{c}| = 6(2\sqrt{2} - 1)$
$(S2): \angle ABC = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$.
Which of the following is true?

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $2 \vec{a}+3 \vec{b}+4 \vec{c}=\vec{0}$. Then $|\vec{b} \times \vec{c}|=$

Let $u = -2 \hat{i} + 2 \hat{j} + \hat{k}$ and $v = \hat{i} - 2 \hat{j} + 2 \hat{k}$. Then the angle between $u$ and $v$ is

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=2$,$|\vec{b}|=3$ and $\vec{a}+t \vec{b}$ and $\vec{a}-t \vec{b}$ are perpendicular,where $t$ is a positive scalar,then

If the vectors $\overline{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\overline{b}=2 \hat{i}+4 \hat{j}+\hat{k}$,and $\overline{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}$ are mutually orthogonal,then $(\lambda, \mu) = $