Let $Q(a,b,c)$ be the image of the point $P(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}.$ Then the distance of $Q$ from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is

If the line $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$,then $4\lambda+9\mu$ is equal to :

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

$ABC$ is a triangle with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $(\lambda^3 + \mu^3 + 5)$ is equal to:

If the lines $\frac{x-k}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$,then the largest possible value of $|\lambda|$ is equal to ..........