Two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect at a point. Then the value of ' $k$ ' is

The length of the perpendicular from the point $P(1,-1,2)$ to the given line $\frac{x+1}{2}=\frac{y-2}{-3}=\frac{z+2}{4}$ is

If $Q(1,0,1)$ is the image of the point $P(a, b, c)$ in the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$ then $a+b+c$ is equal to :

Shortest distance between the lines $\vec{r}=(8+3 \lambda) \hat{\imath}-(9+16 \lambda) \hat{\jmath}+(10+7 \lambda) \hat{k}$ and $\vec{r}=15 \hat{\imath}+29 \hat{\jmath}+5 \hat{k}+\mu(3 \hat{\imath}+8 \hat{\jmath}-5 \hat{k})$ is