Let $a, b, c$ be the position vectors of the vertices of a triangle $ABC$. The vector area of triangle $ABC$ is

Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}$,$\vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c}+\hat{i}) \times (\vec{a}+\vec{b}+\hat{i}) = \vec{a} \times (\vec{c}+\hat{i})$ and $\vec{a} \cdot \vec{c} = -29$. Then $\vec{c} \cdot (-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$. If $\lambda \overrightarrow{a} + 2 \overrightarrow{b}$ and $3 \overrightarrow{a} - \lambda \overrightarrow{b}$ are perpendicular to each other,then the number of values of $\lambda$ in $[-1, 3]$ is:

If the lines $\vec{r} = 2\hat{i} + \hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and $\vec{r} = \hat{i} + \hat{j} - 3\hat{k} + \mu(\hat{j} + 2\hat{k})$ intersect each other,then $(\lambda + \mu)$ is equal to

If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{r}$ is a non-zero vector such that $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$,then $\vec{r} = $

Let $\vec{a} = \lambda \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 4\hat{j} + 4\hat{k}$, and $\vec{c} = \hat{i} + \mu \hat{j} + \hat{k}$. If $\vec{a}$ is parallel to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda - \mu$.