Let $H$ be the orthocentre of an acute angled $\triangle ABC$ and $O$ be its circumcenter. Then,$\vec{HA} + \vec{HB} + \vec{HC}$

Let $\vec{a} = 2\hat{i}-\hat{j}-\hat{k}$,$\vec{b} = 5\hat{i}+\hat{j}-2\hat{k}$,and $\vec{c} = -13\hat{i}-11\hat{j}+4\hat{k}$ be the position vectors of three points $A$,$B$,and $C$ respectively. If $\vec{AB} = \lambda \vec{BC}$ and $\vec{AC} = \mu \vec{CB}$,then find the value of $\lambda + \mu$.

The vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a $\triangle ABC$. The length of the median through $A$ is:

If the points with position vectors $(\alpha \hat{i}+10 \hat{j}+13 \hat{k})$,$(6 \hat{i}+11 \hat{j}+11 \hat{k})$,and $(\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k})$ are collinear,then $(19 \alpha-6 \beta)^2=$

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$, and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

If $\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$,then