$A, B, C$ and $D$, are any four points. If $E$ and $F$ are mid-points of $A C$ and $B D$ respectively, then $\mathbf{A B}+\mathbf{C B}+\mathbf{C D}+\mathbf{A D}=$

The four points whose position vectors are given by $2 a+3 b-c, a-2 b+3 c, 3 a+4 b-2 c$ and $a-6 b+6 c$ are

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 $\frac{x+y+z}{p+q+r}=$