In $\triangle ABC$,the points $P, Q, R$ divide $BC, CA, AB$ in the ratio $3:4, 2:5, 9:5$,respectively,and the point $D$ divides $BC$ in the ratio $2:3$. If $\vec{AP} + \vec{BQ} + \vec{CR} = k \vec{AD}$,then $(14k + 1) : (14k - 1) = $

$p=2 \hat{i}-3 \hat{j}+\hat{k}, q=\hat{i}+\hat{j}-\hat{k}$. If the vectors $a$ and $b$ are the orthogonal projections of $p$ on $q$ and $q$ on $p$ respectively, then $\frac{a \times b}{a \cdot b}=$

If $\overline{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$,$\overline{b}=\hat{i}-2 \hat{j}+\hat{k}$,and $\overline{c}=\hat{i}+\hat{j}-\hat{k}$ are three vectors,and there exists a vector $\overline{r}$ such that $\overline{r} \times \overline{a}=\overline{b}$ and $\overline{r} \cdot \overline{c}=3$,then the value of $|\overline{r}|$ is:

$\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $|\vec{a}+\vec{b}+\vec{c}|=1$ and $\vec{a}$ is perpendicular to $\vec{b}$. If $\vec{c}$ makes angles $\alpha, \beta$ with $\vec{a}, \vec{b}$ respectively, then $\cos \alpha+\cos \beta=$

In quadrilateral $ABCD$,$AB=a$,$BC=b$,$AD=b-a$. If $M$ is the midpoint of $BC$ and $N$ is a point on $DM$ such that $DN=\left(\frac{4}{5}\right) DM$,then $5 AN=$

Let $\theta$ denote the angle between vectors $\vec{a}$ and $\vec{b}$. If $\vec{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$,$\vec{a} \cdot \vec{b}=4$ and $\theta=\cos ^{-1}\left(\frac{4}{21}\right)$,then $\vec{a}+\vec{b}$ is: