If the direction ratios of two lines $L_1$ and $L_2$ are given by $(1, -2, 2)$ and $(-2, 3, -6)$ respectively,then the direction ratios of the line which is perpendicular to the lines $L_1$ and $L_2$ are

$A$ unit vector perpendicular to the plane determined by the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ is:

If the two diagonals of a parallelogram are $\bar{d_1} = \bar{i} + 2\bar{j} + 3\bar{k}$ and $\bar{d_2} = -2\bar{i} + \bar{j} - 2\bar{k}$,then the area of the parallelogram in square units 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 $\frac{\pi}{6}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is

If $\bar{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$,$\bar{b}=\hat{i}-2 \hat{j}-2 \hat{k}$,$\bar{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$ and if $\bar{d}$ is a vector perpendicular to both $\bar{b}$ and $\bar{c}$,and $\bar{a} \cdot \bar{d}=18$,then $|\bar{a} \times \bar{d}|^2=$

Let $\lambda \in R$,$\vec{a} = \lambda \hat{i} + 2 \hat{j} - 3 \hat{k}$,and $\vec{b} = \hat{i} - \lambda \hat{j} + 2 \hat{k}$. If $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$,then $|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|^2$ is equal to