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 at right angles,then $k =$

The equation of the line,passing through $A(1, 2, 3)$ and perpendicular to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} + 3 \hat{j} + 2 \hat{k}$,is

The vector equation of the line $2x+4=3y+1=6z-3$ is

Let $P(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z}{1}$ at a distance $4\sqrt{14}$ from the point $(1, -1, 0)$ and nearer to the origin. Then the shortest distance between the lines $\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}$ and $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$ is equal to

If $\vec{r}=\hat{i}+\hat{j}+t(2 \hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+s(3 \hat{i}-5 \hat{j}+2 \hat{k})$ are the vector equations of two lines $L_1$ and $L_2$, then the shortest distance between them is

The equation of the straight line passing through the points $(4, -5, -2)$ and $(-1, 5, 3)$ is