Let $L$ be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$, whose distance from the point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

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

The equation of the line passing through $(2, 3, 4)$ and parallel to the $Y$-axis is . . . . . . .

The Cartesian equation of a line is $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z-5}{5}$,then the vector equation of the line 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

The lines $\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 3}{0}$ and $\frac{x - 2}{0} = \frac{y - 3}{0} = \frac{z - 4}{1}$ are: