If $a=|\mathbf{a}| ; b=|\mathbf{b}|$, then $\left(\frac{\mathbf{a}}{a^2}-\frac{\mathbf{b}}{b^2}\right)^2$

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three unit vectors such that $x \mathbf{a}+y \mathbf{b}+z \mathbf{c}= p(\mathbf{b} \times \mathbf{c})+q(\mathbf{c} \times \mathbf{a})+r(\mathbf{a} \times \mathbf{b})$. If $(\mathbf{a}, \mathbf{b})=(\mathbf{b}, \mathbf{c})=(\mathbf{c}, \mathbf{a})=\frac{\pi}{3}$, $(\mathbf{a}, \mathbf{b} \times \mathbf{c})=\frac{\pi}{6}$ and $\mathbf{a}, \mathbf{b}, \mathbf{c}$ form a right-handed system, then $\frac{x+y+z}{p+q+r}=$

$O(0,0,0), A(3,1,4), B(1,3,2)$ and $C(0,4,-2)$ are the vertices of a tetrahedron. If $G$ is the centroid of the tetrahedron and $G_1$ is the centroid of its face $A B C$, then the point which divides $G G_1$ in the ratio $1: 2$ is

The position vectors of two points $A$ and $B$ are $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $7 \hat{\mathbf{i}}-\hat{\mathbf{k}}$ respectively. The point $P$ with position vector $-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ is on the line $A B$. If the point $Q$ is the harmonic conjugate of $P$, then the sum of the scalar components of the position vector of $Q$ is