The vector equation of the line $2x+4=3y+1=6z-3$ is

$A$ 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)
$(A) (\frac{7}{3}, \frac{7}{3}, \frac{5}{3})$ $(B) (-1, -1, 0)$ $(C) (1, 1, 1)$ $(D) (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$

$L_1$ is a line passing through the points with position vectors $\hat{i}-2 \hat{j}-\hat{k}$ and $4 \hat{i}-3 \hat{k}$. $L_2$ is a line passing through the points with position vectors $\hat{i}+2 \hat{j}-\hat{k}$ and $2 \hat{i}-4 \hat{j}-5 \hat{k}$. Then the distance between $L_1$ and $L_2$ is

The equation of the straight line $3x + 2y - z - 4 = 0$ and $4x + y - 2z + 3 = 0$ in the symmetrical form is:

The lines whose direction cosines satisfy the equations $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ will be perpendicular if...

Find the vector equation of the line $\frac{x - 2}{2} = \frac{2y - 5}{-3}, z = -1$.