Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{C}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c}=5$, then $|\vec{c}|$ is equal to

Let the point A divide the line segment joining the points $\mathrm{P}(-1,-1,2)$ and $\mathrm{Q}(5,5,10)$ internally in the ratio $r: 1(r>0)$. If O is the origin and $(\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{1}{5}|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=10$, then the value of r is :

Let the position vectors of the vertices $\mathrm{A}, \mathrm{B}$ and C of a tetrahedron ABCD be $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{k}$ and $2 \hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $A B C$ meets the median line segment through $A$ of the triangle $A B C$ at the point $E$. If the length of $A D$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6 \sqrt{2}}$, then the position vector of E is

Let the arc $A C$ of a circle subtend a right angle at the centre $O$. If the point $B$ on the arc $A C$, divides the arc $A C$ such that $\frac{\text { length of } \operatorname{arc} A B}{\text { length of } \operatorname{arc} B C}=\frac{1}{5}$, and $\overrightarrow{O C}=\alpha \overrightarrow{O A}+\beta \overrightarrow{O B}$, then $\alpha+\sqrt{2}(\sqrt{3}-1) \beta$ is equal to