If the line joining points $A$ and $B$ having position vectors $6 \vec{a}-4 \vec{b}+4 \vec{c}$ and $-4 \vec{c}$ respectively,and the line joining the points $C$ and $D$ having position vectors $-\vec{a}-2 \vec{b}-3 \vec{c}$ and $\vec{a}+2 \vec{b}-5 \vec{c}$ intersect,then their point of intersection is

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:

Let $a = i + 2j + k$,$b = i - j + k$,$c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has projection $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is

If $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are unit vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,then the value of $3 \overrightarrow{a} \cdot \overrightarrow{b}+2 \overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}$ is

Find the angle between the pair of lines given by $\vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})$ and $\vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})$.

The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.