Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

Let $a, b, c$ be three distinct real numbers,none equal to $1$. If the vectors $a \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+ c \hat{k}$ are coplanar,then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to

If the vectors $\vec{a} + \lambda \vec{b} + 3\vec{c}$,$-2\vec{a} + 3\vec{b} - 4\vec{c}$,and $\vec{a} - 3\vec{b} + 5\vec{c}$ are coplanar,and $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,find the value of $\lambda$.

$\bar{a}, \bar{b}, \bar{c}$ 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} = $

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

If $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $p, q$ are real numbers,then the equality $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$ holds for: