The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

If $P$ is a point lying on the line passing through the point $A(\hat{i}-\hat{j}+3 \hat{k})$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$ such that $|AP|=18$,then a position vector of $P$ is

Find the vector and the cartesian equations of the lines that passes through the origin and $(5, -2, 3).$

The shortest distance between lines $L_1$ and $L_2$,where $L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and $L_2$ is the line passing through the points $A(-4,4,3)$ and $B(-1,6,3)$ and perpendicular to the line $\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$,is

The point of intersection of the lines $\vec{r}=2 \vec{b}+t(6 \vec{c}-\vec{a})$ and $\vec{r}=\vec{a}+s(\vec{b}-3 \vec{c})$ is

The shortest distance between the line passing through the point $\bar{i} + 2\bar{j} + 3\bar{k}$ and parallel to the vector $2\bar{i} + 3\bar{j} + 4\bar{k}$ and the line passing through the point $2\bar{i} + 4\bar{j} + 5\bar{k}$ and parallel to the vector $3\bar{i} + 4\bar{j} + 5\bar{k}$ is: