If the force $\overrightarrow{F} = \hat{i} + 2\hat{j} + 3\hat{k}$ moves a particle from position $\vec{r_1} = \hat{i} + \hat{j} - \hat{k}$ to $\vec{r_2} = 2\hat{i} - \hat{j} + \hat{k},$ then the work done is:

Let $\lambda \in \mathbb{Z}$,$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0}$,$\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Then $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$ is equal to

If $a = 3i - j + 2k$ and $b = 2i + j - k$,then evaluate $a \times (a \cdot b)$.

In a triangle $ABC,$ right-angled at the vertex $A,$ if the position vectors of $A, B,$ and $C$ are respectively $3\hat{i} + \hat{j} - \hat{k},$ $-\hat{i} + 3\hat{j} + p\hat{k},$ and $5\hat{i} + q\hat{j} - 4\hat{k},$ then the point $(p, q)$ lies on a line

If $\vec{a}$ and $\vec{b}$ are unit vectors,then what is the angle between $\vec{a}$ and $\vec{b}$ for $\sqrt{3}\vec{a} - \vec{b}$ to be a unit vector (in $^{\circ}$)?

Let $ABC$ be an equilateral triangle of side $a$. $M$ and $N$ are two points on the sides $AB$ and $AC$ respectively such that $\overrightarrow{AN} = K \overrightarrow{AC}$ and $\overrightarrow{AB} = 3 \overrightarrow{AM}$. If the vectors $\overrightarrow{BN}$ and $\overrightarrow{CM}$ are perpendicular,then $K =$