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

The unit vectors orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and coplanar with $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ are

$a, b, c, d$ are coplanar vectors, then $(a \times b) \times (c \times d)$ is equal to

The area of the triangle with vertices $(1,2,0)$,$(1,0,2)$ and $(0,3,1)$ is.

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 $60^{\circ}$,then the value of $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is

If $a = 2i + 4j - 5k$ and $b = i + 2j + 3k$,then $|a \times b|$ is