Consider a $\triangle ABC$ where $A(1,3,2)$,$B(-2,8,0)$,and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$,then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $|(\hat{a}+\hat{b})+2(\hat{a} \times \hat{b})|=2$. If $\theta \in(0, \pi)$ is the angle between $\hat{a}$ and $\hat{b}$,then among the statements:
$(S_{1})$: $2|\hat{a} \times \hat{b}|=|\hat{a}-\hat{b}|$
$(S_{2})$: The projection of $\hat{a}$ on $(\hat{a}+\hat{b})$ is $\frac{1}{2}$

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $|\vec{a}+3 \vec{b}-4 \vec{c}|^2=$

The angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ is

The locus of a point equidistant from two given points $a$ and $b$ is given by

Let $a = i + 2j + k$,$b = i - j + k$,$c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has projection $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is