If $a$ and $b$ are unit vectors perpendicular to each other, then $[a + (a \times b) \quad b + (a \times b) \quad (a \times b)]$ =

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to:

$\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $x \bar{a} + y \bar{b} + z \bar{c} = p(\bar{b} \times \bar{c}) + q(\bar{c} \times \bar{a}) + r(\bar{a} \times \bar{b})$. If $(\bar{a}, \bar{b}) = (\bar{b}, \bar{c}) = (\bar{c}, \bar{a}) = \frac{\pi}{3}$, $(\bar{a}, \bar{b} \times \bar{c}) = \frac{\pi}{6}$ and $\bar{a}, \bar{b}, \bar{c}$ form a right-handed system, then $\frac{x+y+z}{p+q+r} = $

If the vectors $2 \hat{i}+3 \hat{j}+4 \hat{k}$,$2 \hat{i}+\hat{j}-\hat{k}$ and $\lambda \hat{i}-\hat{j}+2 \hat{k}$ are coplanar,then the value of $\lambda$ is:

Let $\overrightarrow{OP} = \frac{\alpha-1}{\alpha} \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{OQ} = \hat{i} + \frac{\beta-1}{\beta} \hat{j} + \hat{k}$ and $\overrightarrow{OR} = \hat{i} + \hat{j} + \frac{1}{2} \hat{k}$ be three vectors,where $\alpha, \beta \in \mathbb{R} - \{0\}$ and $O$ denotes the origin. If $(\overrightarrow{OP} \times \overrightarrow{OQ}) \cdot \overrightarrow{OR} = 0$ and the point $(\alpha, \beta, 2)$ lies on the plane $3x + 3y - z + l = 0$,then the value of $l$ is:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}+\hat{k}$, $\vec{c}=\hat{i}+3\hat{j}-2\hat{k}$, and $\vec{d}=2\hat{i}+\hat{j}-\hat{k}$ be four vectors. Let $l=\vec{b} \cdot \vec{c}$ and $m=\vec{b} \cdot \vec{a}$. Find the value of the scalar triple product $[(m\vec{b}+l\vec{a}) \quad \vec{b} \quad \vec{d}]$.