The magnitude of vectors $\vec{a}$ and $\vec{b}$ are $1$ and $2$ respectively,and $\vec{a} \cdot \vec{b} = 1$. Then,the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .

$A$ vector $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$ (where $\alpha, \beta \in R$) lies in the plane of the vectors $\overrightarrow{b} = \hat{i} + \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} + 4 \hat{k}$. If $\overrightarrow{a}$ bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$,then:

If $12 \hat{i}-12 \hat{j}-18 \hat{k}$,$-3 \hat{i}-6 \hat{j}-9 \hat{k}$ and $3 \hat{i}+3 \hat{j}-24 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle ABC$,then the position vector of the incentre of $\triangle ABC$ is

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 $ABC$ be a triangle and $P$ be a point inside $ABC$ such that $\overrightarrow{PA} + 2\overrightarrow{PB} + 3\overrightarrow{PC} = \vec{0}$. The ratio of the area of $\triangle ABC$ to that of $\triangle APC$ is

If $A(3,2,3)$, $B(1,4,6)$, and $C(7,4,5)$ are the three vertices of a parallelogram $ABCD$, then the angle between its diagonal through $D$ and the side $DC$ is