If the vectors $3i + 2j + 8k$ and $2i + xj + k$ are perpendicular to each other,then $x = \dots$

The cosine of the angle $A$ of the triangle with vertices $A(1, -1, 2)$,$B(6, 11, 2)$,and $C(1, 2, 6)$ is:

Let $\vec{p}$ and $\vec{q}$ be the position vectors of points $P$ and $Q$ respectively,with respect to the origin $O$,and let $|\vec{p}|=p, |\vec{q}|=q$. The points $R$ and $S$ divide the line segment $PQ$ internally and externally in the ratio $2:3$ respectively. If $\vec{OR}$ and $\vec{OS}$ are perpendicular,then:

Let $A = \hat{i} + 2 \hat{j}$. If $B$ is a vector in the $XY$ plane such that $(A + B) \cdot B = 15$ and $A \cdot B = 6$,then $|B|$ is

$A$ particle acted on by two forces $3i + 2j - 3k$ and $2i + 4j + 2k$ is displaced from the point $i + 2j + k$ to $5i + 4j + 2k.$ The total work done by the forces is equal to ............ $unit$.

The position vectors of the vertices of $\triangle ABC$ are $4\hat{i} - 2\hat{j}$,$\hat{i} + 4\hat{j} - 3\hat{k}$,and $-\hat{i} + 5\hat{j} + \hat{k}$ respectively. Then,$m \angle ABC = $