If $a = 3i - j + 2k$ and $b = 2i + j - k$,then evaluate $a \times (a \cdot b)$.

The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i}-3 \hat{j}+3 \hat{k}$,$2 \hat{i}+2 \hat{j}+3 \hat{k}$ and $-\hat{i}+\hat{j}+3 \hat{k}$ respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$,where $D$ is on the line segment $BC$. Then $2 l^2$ equals:

If $\theta$ is the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$,then $\cos \frac{\theta}{2} = $

Statement $(A):$ If $|\vec{a}| = 2, |\vec{b}| = 3, |2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}| = 5$.
Reason $(R): |\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$

Let the angle $\theta, 0 < \theta < \frac{\pi}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin^{-1}\left(\frac{\sqrt{65}}{9}\right)$. If the vector $\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$,then the value of $9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$ is

The vector projection of $\vec{b}$ on $\vec{a}$ where $\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and $\vec{b}=7 \hat{i}-5 \hat{j}-\hat{k}$ is: