If the area of the triangle with vertices $(1, 2, 0)$,$(1, 0, 2)$ and $(0, x, 1)$ is $\sqrt{6}$ square units,then the value of $x$ is

Let $\overrightarrow{c}$ be a vector perpendicular to the vectors $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}.$ If $\overrightarrow{c}\cdot(\hat{i}+\hat{j}+3\hat{k})=8,$ then the value of $\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b})$ is equal to ...... .

Let $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is :

For any vector $x$, where $\hat{i}, \hat{j}, \hat{k}$ have their usual meanings, the value of $(x \times \hat{i})^{2} + (x \times \hat{j})^{2} + (x \times \hat{k})^{2}$ is equal to

$a$ and $b$ are two vectors such that $|a|=\sqrt{3}$ and $|b|=\sqrt{2}$. If $x$ is a unit vector satisfying $x \times a = b$,then $x$ is equal to:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+\hat{k}$ are two vectors,and $\vec{c}$ is a unit vector lying in the plane of $\vec{a}$ and $\vec{b}$ such that $\vec{c}$ is perpendicular to $\vec{b}$,then find the value of $\vec{c} \cdot (\hat{i}+\hat{j}+2\hat{k})$.