$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non- coplanar and mutually perpendicular vectors of same magnitude $K . r$ is any vectors satisfying $\mathbf{a} \times((\mathbf{r}-\mathbf{b}) \times \mathbf{a})+\mathbf{b} \times((\mathbf{r}-\mathbf{c}) \times \mathbf{b})+\mathbf{c} \times((\mathbf{r}-\mathbf{a}) \times \mathbf{c})=\mathbf{0}$, then $\mathbf{r}=$

Consider the following

Assertion (A) The two lines $\mathbf{r}=\mathbf{a}+t(\mathbf{b})$ and $\mathbf{r}=\mathbf{b}+s(\mathbf{a})$ intersect each other.

Reason (R) The shortest distance between the lines $\mathbf{r}=\mathbf{p}+t(\mathbf{q})$ and $\mathbf{r}=\mathbf{c}+s(\mathbf{d})$ is equal to the length of projection of the vector ( $\mathbf{p}-\mathbf{c}$ ) on ( $\mathbf{q} \times \mathbf{d}$ )

The correct answer is

$A B C D$ is a tetrahedron, $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}},-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$, $3 \bar{i}+2 \bar{j}-\bar{k}$ are the the position vectors of the points $A, B$ and $C$ respectively. $-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ is the position vector of the centroid of the triangular face $B C D$. If G is the centroid of the tetrahedron, then $G D=$

If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{c}=-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}$ are three vectors, then $\sqrt{(|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|)+|\mathbf{a}+\mathbf{b}+\mathbf{c}|}=$