If the points with position vectors $\hat{i}-2 \hat{j}+3 \hat{k}$, $2 \hat{i}+3 \hat{j}-4 \hat{k}$, $-3 \hat{i}+\hat{j}-5 \hat{k}$, and $a \hat{i}-2 \hat{j}+4 \hat{k}$ are coplanar, then $a=$

If $a, b, c$ are three non-coplanar vectors and $d$ is any unit vector, then $|(a \cdot d)(b \times c) + (b \cdot d)(c \times a) + (c \cdot d)(a \times b)| = $

If $\bar{a} = \bar{i} - \bar{j}$,$\bar{b} = \bar{j} - \bar{k}$,$\bar{c} = \bar{k} - \bar{i}$ and $\bar{d}$ is a unit vector such that $\bar{a} \cdot \bar{d} = 0$ and $[\bar{b} \bar{c} \bar{d}] = 0$,then the vector $\bar{d} = ....$

If $a = 3i - 2j + 2k$,$b = 6i + 4j - 2k$ and $c = 3i - 2j - 4k$,then $a \cdot (b \times c)$ is

Let $\vec{\lambda} = x\vec{a} + y\vec{b} + z\vec{c}$ and $\vec{\lambda} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = 2(x + y + z)$ (where $x + y + z \neq 0$),then the scalar triple product $[\vec{a} \, \vec{b} \, \vec{c}]$ is:

$\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar vectors such that $\overrightarrow{P} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}$,$\overrightarrow{Q} = 4\overrightarrow{a} + 3\overrightarrow{b} + 4\overrightarrow{c}$,and $\overrightarrow{R} = \overrightarrow{a} + \alpha\overrightarrow{b} + \beta\overrightarrow{c}$ are linearly dependent vectors. Then,the number of possible values of $\alpha$ is: