If $a+xb+yc=0$ and $a \times b+b \times c+c \times a=6(b \times c)$,then the locus of the point $(x, y)$ is

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. Let $\overline{c}$ be a vector such that $|\bar{c}-\bar{a}|=3$ and $|(\bar{a} \times \bar{b}) \times \bar{c}|=3$ and the angle between $\overline{c}$ and $\overline{a} \times \overline{b}$ is $30^{\circ}$,then $\overline{a} \cdot \overline{c}$ is equal to

The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$ is

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

If $\vec{a}$ and $\vec{b}$ are unit vectors such that the vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$,then find the angle between $\vec{a}$ and $\vec{b}$.

If the constant forces $2 \hat{i}-5 \hat{j}+6 \hat{k}$ and $-\hat{i}+2 \hat{j}-\hat{k}$ act on a particle due to which it is displaced from a point $A(4,-3,-2)$ to a point $B(6,1,-3)$, then the work done by the forces is (in $\text{ unit}$)