The equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ 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)
$(A) (\frac{7}{3}, \frac{7}{3}, \frac{5}{3})$ $(B) (-1, -1, 0)$ $(C) (1, 1, 1)$ $(D) (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

The distance from the point $-i + 2j + 6k$ to the straight line passing through the point $(2, 3, -4)$ and parallel to the vector $6i + 3j - 4k$ is

$L_1$ 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 line $L_1$ is parallel to the vector $\vec{a} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$,and the line $L_2$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is: