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

For non-zero vectors $\bar{a}, \bar{b}, \bar{c}$,if $\bar{a} \times \bar{b} = \bar{c}$ and $\bar{b} \times \bar{c} = \bar{a}$,then:

Let the lines $L_1: \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}$ and $L_2: \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$. The unit vector perpendicular to both $L_1$ and $L_2$ is:

For any vector $\vec{a} \in \mathbb{R}^3$,$|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = $ . . . . . . .

If $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}$,$|\vec{b}| = 6$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then the area of the triangle (in square units) with $\vec{a}$ and $\vec{b}$ as two of its sides is

For any two vectors $a$ and $b$,if $a \times b = 0$,then