$A$ 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:

Let $(x, y) \in (R \times R)$ and $\vec{a} = x \hat{i} + 2 \hat{j} - \hat{k}$, $\vec{b} = 6 \hat{i} - y \hat{j} + 2 \hat{k}$ be two vectors. If $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = f(x) g(y)$, then $f(x) + g(y) - 46 = 0$ represents:

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to :

Let $\theta$ denote the angle between vectors $\vec{a}$ and $\vec{b}$. If $\vec{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$,$\vec{a} \cdot \vec{b}=4$ and $\theta=\cos ^{-1}\left(\frac{4}{21}\right)$,then $\vec{a}+\vec{b}$ is:

If $\bar{a}, \bar{b}, \bar{c}$ are perpendicular to $\bar{b}+\bar{c}, \bar{c}+\bar{a}$ and $\bar{a}+\bar{b}$ respectively and $|\bar{a}+\bar{b}|=2, |\bar{b}+\bar{c}|=6, |\bar{c}+\bar{a}|=4$,then $|\bar{a}+\bar{b}+\bar{c}|=$

If $l\vec{a} + m\vec{b} + n\vec{c} = \vec{0},$ where $l, m, n$ are scalars and $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular non-zero vectors,then