If the vectors $a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}, \hat{i}+\hat{j}+c \hat{k}$ $(a \neq 1, b \neq 1, c \neq 1)$ are coplanar,then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is

If $A$,$B$,and $C$ are three non-coplanar vectors,then $(A + B + C) \cdot ((A + B) \times (A + C)) = \dots$

$(\vec{a}+2 \vec{b}-\vec{c}) \cdot \{(\vec{a}-\vec{b}) \times (\vec{a}-\vec{b}-\vec{c})\} =$

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$,$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$,and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$ be three non-zero vectors such that $\vec{c}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. If the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2 = \dots$

The vector $(\hat{i} \times \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} \times \vec{a} \cdot \vec{b})\hat{j} + (\hat{k} \times \vec{a} \cdot \vec{b})\hat{k}$ is equal to

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = x\hat{i} + (x - 2)\hat{j} - \hat{k}$ are three vectors such that $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, then $x = ...$