The distance of the point $(2, 3, 4)$ from the line $1 - x = \frac{y}{2} = \frac{1}{3}(1 + z)$ is

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is

If $L_1$ is a line through the point $5 \hat{i}+8 \hat{j}+11 \hat{k}$ and parallel to the vector $2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $L_2$ is a line through the point $4 \hat{i}+6 \hat{j}+8 \hat{k}$ and parallel to the vector $3 \hat{i}+4 \hat{j}+5 \hat{k}$, then the point of intersection of $L_1$ and $L_2$ is

If $\frac{x - 1}{l} = \frac{y - 2}{m} = \frac{z + 1}{n}$ is the equation of the line passing through the points $(1, 2, -1)$ and $(-1, 0, 1)$,then the direction ratios $(l, m, n)$ are:

The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$

If a point $R(4, y, z)$ lies on the line joining the points $P(2, -3, 4)$ and $Q(8, 0, 10)$,then the distance of $R$ from the origin is