If the 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,then $k =$

The orthocentre of the triangle formed by the points $(2,1,5)$,$(3,2,3)$,and $(4,0,4)$ is

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line $\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}$ and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$,then $K$ is equal to:

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 ..........

If the shortest distance between the lines
$L_1: \overrightarrow{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, \lambda \in R$
$L_2: \overrightarrow{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \mu \in R$
is $\frac{m}{\sqrt{n}}$,where $\operatorname{gcd}(m, n)=1$,then the value of $m+n$ equals.

The cosine of the angle included between the lines $\overline{r}=(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}-2 \hat{k})$ and $\overline{r}=(\hat{\imath}+\hat{\jmath}+3 \hat{k})+\mu(3 \hat{\imath}+2 \hat{\jmath}-6 \hat{k})$ where $\lambda, \mu \in R$ is