Let $p_1, p_2$ and $p_3$ be the altitudes of a $\triangle A B C$ drawn through the vertices $A, B$ and $C$ respectively. If $r_1=4$, $r_2=6, r_3=12$ are the ex-radii of $\triangle A B C$, then $\frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}=$

$A B C D$ 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