$P, Q, R$ and $S$ are four points with the position vectors $3i-4j+5k, 0i+0j+4k, -4i+5j+1k$ and $-3i+4j+3k$,respectively. Then,the line $PQ$ meets the line $RS$ at the point

The equation of the line passing through the points $(3, 4, -7)$ and $(6, -1, 1)$ is

The Cartesian equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

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 acute angle between the lines $x=-y, z=0$ and $x=0, z=0$ is

The coordinates of the points on the line $\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$ at a distance of $12 \text{ units}$ from the point $A(-2, 1, -1)$ are