$A$ force of magnitude $5$ units acting along the vector $2i - 2j + k$ displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$. The work done is:

$ABC$ is a right-angled triangle in which $\max \{AB, BC, AC\} = BC$. If the position vectors of $B$ and $C$ are respectively $3\hat{i}-2\hat{j}+\hat{k}$ and $5\hat{i}+\hat{j}-3\hat{k}$,then find the value of $AB \cdot AC + BA \cdot BC + CA \cdot CB$.

If the vectors $\bar{a}=\hat{i}-\hat{j}+2 \hat{k}$,$\bar{b}=2 \hat{i}+4 \hat{j}+\hat{k}$,and $\bar{c}=p \hat{i}+\hat{j}+q \hat{k}$ are mutually orthogonal,then $(p, q)$ is equal to

If the vector $a + b$ makes equal angles with vectors $a$ and $b$,then:

If $A(4,7,8)$,$B(2,3,4)$,and $C(2,5,7)$ are the position vectors of the vertices of a triangle $ABC$ and if the internal bisector of $\angle A$ meets $BC$ at $D$,then $AD=$

Two adjacent sides of a parallelogram are represented by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $\vec{AD}$ is rotated by an acute angle $\theta$ in the plane of the parallelogram such that $\vec{AD}$ becomes $\vec{AD'}$. If $\vec{AD'}$ is perpendicular to $\vec{AB}$,find $\cos \theta$.