If $a=2\hat{i}+\hat{j}-3\hat{k}$,$b=\hat{i}-2\hat{j}+\hat{k}$,$c=-\hat{i}+\hat{j}-4\hat{k}$ and $d=\hat{i}+\hat{j}+\hat{k}$,then $|(a \times b) \times(c \times d)|=$

If $a = (1, -1, 1)$ and $c = (-1, -1, 0),$ then the vector $b$ satisfying $a \times b = c$ and $a \cdot b = 1$ is

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k,$ then a unit vector perpendicular to both $u$ and $v$ is

Direction ratios of the line which is perpendicular to the lines with direction ratios $-1, 2, 2$ and $0, 2, 1$ are

Let $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$,$\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c}=\vec{c} \times \vec{b}$ and $(\overrightarrow{a}+\overrightarrow{c}) \cdot(\overrightarrow{b}+\overrightarrow{c})=168$. Then the maximum value of $|\vec{c}|^2$ is :

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