Find the equation of a line parallel to the $x$-axis and passing through the origin.

$A$ line $l$ passing through the origin is perpendicular to the lines
$l_1: (3+t) \hat{i} + (-1+2t) \hat{j} + (4+2t) \hat{k}, -\infty < t < \infty$
$l_2: (3+2s) \hat{i} + (3+2s) \hat{j} + (2+s) \hat{k}, -\infty < s < \infty$
Then,the coordinate$(s)$ of the point$(s)$ on $l_2$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_1$ is(are)
$(A) (\frac{7}{3}, \frac{7}{3}, \frac{5}{3})$ $(B) (-1, -1, 0)$ $(C) (1, 1, 1)$ $(D) (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$

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 $B(1, 6, 3)$.

Show that the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}$ intersect. Also,find their point of intersection.

The equation of a line passing through the point $(2, -1, 1)$ and parallel to the line whose equation is $\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{-3}$ is:

Statement-$1$: The shortest distance between the skew lines $\frac{x+3}{-4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{x+3}{-4} = \frac{y}{1} = \frac{z-7}{1}$ is $9$.
Statement-$2$: Two lines are skew lines if there exists no plane passing through them.