If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then the value of $k$ 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 angle between the two lines $\frac{x-3}{1}=\frac{y-2}{2}=\frac{z+4}{2}$ and $\frac{x-5}{3}=\frac{y+2}{2}=\frac{z}{6}$ is $\qquad$ .

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

Let the point,on the line passing through the points $P(1, -2, 3)$ and $Q(5, -4, 7)$,farther from the origin and at a distance of $9$ units from the point $P$,be $(\alpha, \beta, \gamma)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

Find the vector equation for the line passing through the points $(-1, 0, 2)$ and $(3, 4, 6).$