If $\bar{a}=\hat{i}+\hat{j}$ and $\bar{b}=2 \hat{i}-\hat{k}$,then the point of intersection of the lines $\bar{r} \times \bar{a}=\bar{b} \times \bar{a}$ and $\bar{r} \times \bar{b}=\bar{a} \times \bar{b}$ is

Let $\bar{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}$,$\bar{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k}$ and $\overline{c}=\hat{i}+2 \hat{j}-2 \hat{k}$,where $\alpha, \beta \in R$,be three vectors. If the projection of $\overline{a}$ on $\overline{c}$ is $\frac{10}{3}$ and $\bar{b} \times \bar{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}$,then the value of $2 \alpha+\beta$ is

Given $a = i + j - k$,$b = -i + 2j + k$,and $c = -i + 2j - k$. $A$ unit vector perpendicular to both $a + b$ and $b + c$ is

Let $L_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1}$ and $L_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $L_1$ and $L_2$ is

If $|a|=1, |b|=2$ and the angle between $a$ and $b$ is $120^{\circ}$, then ${(a+3b) \times (3a-b)}^2$ is equal to

$A$ force $\overrightarrow{F} = 2i + j - k$ acts at a point $A$,whose position vector is $2i - j$. The moment of $\overrightarrow{F}$ about the origin is: