If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are perpendicular to each other,then $k$ is

Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is

If the line passing through the points $(5, 1, a)$ and $(3, b, 1)$ crosses the $YZ$ plane at the point $(0, 17/2, -13/2)$,then $a+b=$

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

The point of intersection of the lines represented by $r=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ and $r=(-\hat{i}-3 \hat{j}+7 \hat{k})+\mu(\hat{i}+2 \hat{j}-\hat{k})$ is

The angle between the lines $\vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k})+\lambda(\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ is