The shortest distance between the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$ is

Let $A, B, C$ be three points whose position vectors respectively are:

$\overrightarrow{a}=\hat{i}+4\hat{j}+3\hat{k}$

$\overrightarrow{b}=2\hat{i}+\alpha \hat{j}+4\hat{k}, \alpha \in R$

$\overrightarrow{c}=3\hat{i}-2\hat{j}+5\hat{k}$

If $\alpha$ is the smallest positive integer for which $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-collinear, then the length of the median of $△ABC$ through $A$ is:

Let $ABC$ be a triangle such that $\overrightarrow{BC}=\overrightarrow{\mathit{a}},\overrightarrow{CA}=\overrightarrow{b}$, $\overrightarrow{AB}=\overrightarrow{\mathit{c}},\left|\overrightarrow{a}\right|=6\sqrt{2},\left|\overrightarrow{b}\right|=2\sqrt{3}$ and $\overrightarrow{b}·\overrightarrow{\mathit{c}}=12$

Consider the statements :

$\left(S1\right)$: $\left|\left(\overrightarrow{a}\times \overrightarrow{\mathit{b}}\right)+\left(\overrightarrow{c}\times \overrightarrow{\mathit{b}}\right)\right|-\left|\overrightarrow{c}\right|=6\left(2\sqrt{2}-1\right)$

$\left(S2\right)$: $\angle ACB={\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)$. Then