The foot of the perpendicular from $$(2,4,-1)$$ to the line $$x+5=\frac{1}{4}(y+3)=-\frac{1}{9}(z-6)$$ is

$$P$$ is a point on the line segment joining the points $$(3,2,-1)$$ and $$(6,2,-2)$$. If $$x$$ coordinate of $$\mathrm{P}$$ is 5, then its $$y$$ co-ordinate is

The vector equation of two lines are

$$\begin{aligned} & \vec{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k} \\ & \vec{r}=(s+1) \hat{\imath}+(2 s-1) \hat{\jmath}-(2 s+1) \hat{k} \end{aligned}$$

Then the shortest distance between them is

The place $$x-2 y+z=0$$ is parallel to the line