The sum of the coordinates of one of the points on the line $\frac{x - 2}{1} = \frac{y + 3}{-2} = \frac{z + 5}{2}$ which is at a distance of $3 \text{ units}$ from the point $(2, -3, -5)$ is

The angle between two lines $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}$ and $\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}$ is

The vector equation of the line whose Cartesian equations are $y=2$ and $4x-3z+5=0$ is

The parametric equations of the line passing through the points $A(3,4,-7)$ and $B(1,-1,6)$ are

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.

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$,and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular drawn from $A$ to $BC$,then the coordinates of $D$ are