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

Vectors $\bar{a}$ and $\bar{b}$ are such that $|\bar{a}|=1$,$|\bar{b}|=4$ and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=2 \bar{a} \times \bar{b}-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is

If $a + b + c = 0,$ then which relation is correct?

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}$ and if $6 \hat{i}+2 \hat{j}+3 \hat{k}=\lambda_1(\vec{a} \times \vec{b})+\lambda_2(\vec{b} \times \vec{c})+\lambda_3(\vec{c} \times \vec{a})$,then $(\lambda_1, \lambda_2, \lambda_3)=$

Let $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k}$ and $\overrightarrow{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k}$,where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15(\alpha^{2} + 4)}$,then the value of $2|\vec{a}|^{2} + (\vec{a} \cdot \vec{b})|\vec{b}|^{2}$ is equal to

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