Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$. Then $|\vec{a}+2\vec{b}|^{2}+|\vec{a}+2\vec{c}|^{2}$ is equal to

If $\theta$ is the angle between the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$ and $a \hat{i}+4 \hat{j}+b \hat{k}$ and $\cos \theta=\frac{2}{3}$, then $2(a+b+3)=$

If $\overline{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\overline{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\overline{c}=3 \hat{i}+\hat{j}$ such that $\overline{b}+\lambda \overline{a}$ is perpendicular to $\overline{c}$,then $\lambda$ is

If $\vec{\lambda}$ is a unit vector perpendicular to the plane of vectors $\vec{a}$ and $\vec{b}$,and the angle between them is $\theta$,then $\vec{a} \cdot \vec{b}$ will be:

If $|a \times b|^2 + |a \cdot b|^2 = 36$ and $|a| = 3$,then $|b|$ is equal to

If $A(4,7,8)$,$B(2,3,4)$,and $C(2,5,7)$ are the position vectors of the vertices of a triangle $ABC$ and if the internal bisector of $\angle A$ meets $BC$ at $D$,then $AD=$