$A$ unit vector which is perpendicular to the vector $2\hat{i} - \hat{j} + 2\hat{k}$ and is coplanar with the vectors $\hat{i} + \hat{j} - \hat{k}$ and $2\hat{i} + 2\hat{j} - \hat{k}$ is

Let $A(2, 3, 0)$, $B(0, 3, 2)$ and $C(4, 0, 3)$ be the vertices of a triangle. Find the area of the triangle.

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$,and the angle between $\overline{a} \times \overline{b}$ and $\overline{c}$ is $\frac{2 \pi}{3}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|=$

If $a=2 \hat{i}+3 \hat{j}-5 \hat{k}$,$b=m \hat{i}+n \hat{j}+12 \hat{k}$ and $a \times b=0$,then $(m, n)$ is equal to

$x, y, z$ are three vectors each of magnitude $\sqrt{2}$ and each making an angle $60^{\circ}$ with one another. If $a=x \times(y \times z), b=y \times(z \times x)$, $c=x \times y$, then $x=$

Let $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{v}=-3 \hat{j}+2 \hat{k}$ be vectors in $R^3$ and $\vec{w}$ be a unit vector in the $XY$-plane. Then,the maximum value of $|(\vec{u} \times \vec{v}) \cdot \vec{w}|$ is: