The moment of a force represented by $\overrightarrow{F} = i + 2j + 3k$ about the point $P(2, -1, 1)$ is given by the cross product of the position vector $\overrightarrow{r}$ and the force vector $\overrightarrow{F}$. If the origin is $O(0, 0, 0)$,then $\overrightarrow{r} = 2i - j + k$. Calculate the moment $\overrightarrow{M} = \overrightarrow{r} \times \overrightarrow{F}$.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k},$ find a vector $\vec{c}$ such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3.$

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,$\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$,then $\lambda+\mu$ is equal to:

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{3}$,$|\vec{b}|=5$,$\vec{b} \cdot \vec{c}=10$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$. If $\vec{a}$ is perpendicular to the vector $\vec{b} \times \vec{c}$,then $|\vec{a} \times (\vec{b} \times \vec{c})|$ is equal to:

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$,$\overrightarrow{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$,$\overrightarrow{c} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\overrightarrow{d}$ be a vector such that $\overrightarrow{b} \times \overrightarrow{d} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \cdot \overrightarrow{d} = 4$. Then $|(\overrightarrow{a} \times \overrightarrow{d})|^2$ is equal to . . . . . . .

The direction ratios of the line perpendicular to the lines having direction ratios $2, 3, 1$ and $1, 2, 1$ are