Let $ABC$ be a triangle. Let a point $P$ divide $AB$ in the ratio $1:2$ internally and a point $Q$ divide $BC$ in the ratio $1:2$ internally. Let $D$ be the point of intersection of $AQ$ and $CP$. If the area of the triangle $ABC$ is $k$ square units,then the area of the triangle $BCD$ in square units is:

If $\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}-3 \hat{j}+\hat{k}$ and the orthogonal projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$,then $\lambda$ is equal to

Let $\overline{a}, \overline{b}$ and $\overline{c}$ be vectors of magnitude $2, 3$ and $4$ respectively. If $\overline{a}$ is perpendicular to $(\overline{b}+\overline{c})$,$\overline{b}$ is perpendicular to $(\overline{c}+\overline{a})$ and $\overline{c}$ is perpendicular to $(\overline{a}+\overline{b})$,then the magnitude of $\overline{a}+\overline{b}+\overline{c}$ is equal to

The vector $\vec{a} = \alpha \hat{i} + 2\hat{j} + \beta \hat{k}$ lies in the plane of $\vec{b} = \hat{i} + \hat{j}$ and $\vec{c} = \hat{j} + \hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Find the possible values of $\alpha$ and $\beta$.

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

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