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:

If the line joining the points $(k, 3, 4)$ and $(4, 7, 8)$ is parallel to the line joining the points $(-1, -2, 1)$ and $(1, 2, l)$,then $k + l =$

The point collinear with $(1, -2, -3)$ and $(2, 0, 0)$ among the following is

Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.
Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

Find the vector and the Cartesian equations of the line passing through the point $(5, 2, -4)$ and parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?