Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a\hat{i} + a\hat{j} + c\hat{k}$,$\hat{i} + \hat{k}$ and $c\hat{i} + c\hat{j} + b\hat{k}$ are coplanar,then $c = \dots$

If $a, b, c$ are three non-coplanar vectors,then $\frac{a \cdot (b \times c)}{c \times a \cdot b} + \frac{b \cdot (a \times c)}{c \cdot (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:

If $a, b, c$ are any three non-coplanar vectors,then $[a + b, b + c, c + a] = $

If $\overline{a}=\hat{i}-\hat{k}$,$\overline{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\overline{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$,then $\overline{a} \cdot(\overline{b} \times \overline{c})$ depends on

The scalar $\overline{a} \cdot [(\overline{b} + \overline{c}) \times (\overline{a} + \overline{b} + \overline{c})]$ equals