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 $\bar{a} = 2\hat{i} + \hat{j} + \hat{k}$,$\bar{b} = \hat{i} + 2\hat{j} - \hat{k}$,and vector $\bar{c}$ be coplanar with $\bar{a}$ and $\bar{b}$. If $\bar{c}$ is perpendicular to $\bar{a}$,then $\bar{c}$ is:

The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is

The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the line $\vec{r} = 3\hat{i} - \hat{j} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ is:

$\bar{a}, \bar{b}, \bar{c}$ are three non-coplanar and mutually perpendicular vectors of same magnitude $K$. If $\bar{r}$ is any vector satisfying $\bar{a} \times ((\bar{r}-\bar{b}) \times \bar{a}) + \bar{b} \times ((\bar{r}-\bar{c}) \times \bar{b}) + \bar{c} \times ((\bar{r}-\bar{a}) \times \bar{c}) = \bar{0}$, then $\bar{r} =$

In a triangle $PQR$,let $\vec{a}=\vec{QR}, \vec{b}=\vec{RP}$ and $\vec{c}=\vec{PQ}$. If $|\vec{a}|=3, |\vec{b}|=4$ and $\frac{\vec{a} \cdot(\vec{c}-\vec{b})}{\vec{c} \cdot(\vec{a}-\vec{b})}=\frac{|\vec{a}|}{|\vec{a}|+|\vec{b}|}$,then the value of $|\vec{a} \times \vec{b}|^2$ is.