If for some $m \in \mathbb{R}$ the lines $L_1 : \frac{x + 1}{m} = \frac{y - m}{-1} = \frac{z - 1}{1}$ and $L_2 : \frac{x + 2}{-4} = \frac{y + 1}{9} = \frac{z + 1}{1}$ are coplanar, then line $L_1$ passes through the point

If the lines $2x = ky = -z$ and $6x = -y = -4z$ are perpendicular to each other, then the value of $k$ is:

The angle between the pair of lines given by $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} - 2\hat{k} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$ is . . . . . . .

The coordinates of the point where the line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ is intersected by the $YOZ$ plane are

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

If $(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from a point $P(-1, 2, -1)$ to the line joining the points $A(2, -1, 1)$ and $B(1, 1, -2)$, then $\alpha + \beta + \gamma =$