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 $\sqrt{(|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}|)+|\mathbf{a}+\mathbf{b}+\mathbf{c}|}=$

Let $\mathbf{a}$ and $\mathbf{b}$ be two vectors such that $|\mathbf{a}|=|\mathbf{b}|$ and $|\mathbf{a}+2 \mathbf{b}|=|2 \mathbf{a}-\mathbf{b}|$. If $\mathbf{c}$ is a vector parallel to $\mathbf{a}$, then the angle between $\mathbf{b}$ and $\mathbf{c}$ is