$A$ unit vector in the plane of the vectors $2i + j + k$ and $i - j + k$ and orthogonal to $5i + 2j + 6k$ is

Two forces $\vec{F_1} = 2\hat{i} - 5\hat{j} + 6\hat{k}$ and $\vec{F_2} = -\hat{i} + 2\hat{j} - \hat{k}$ act on a particle. The particle is displaced from point $P(4\hat{i} - 3\hat{j} + 2\hat{k})$ to point $Q(6\hat{i} + \hat{j} + 3\hat{k})$. The work done by the forces is ............. units.

If $\theta$ is the angle between the vectors $\bar{a}$ and $\bar{b}$ where $|\bar{a}|=4, |\bar{b}|=3$ and $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{3}\right)$,then $|(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a} \cdot \bar{b})^2$ has the value

If the vectors $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$,$\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$,and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ form a $\triangle ABC$,then the values of $p, q, r$ and $s$ such that the area of that $\triangle ABC$ is $5 \sqrt{6}$ are:

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=2$,$|\vec{b}|=3$ and $\vec{a}+t \vec{b}$ and $\vec{a}-t \vec{b}$ are perpendicular,where $t$ is a positive scalar,then

If the vectors $a\,i - 2j + 3k$ and $3i + 6j - 5k$ are perpendicular to each other,then $a$ is given by