If the line joining the points $(k, 3, 4)$ and $(4, 7, 8)$ is parallel to the line joining the points $(-1, -2, 1)$ and $(1, 2, l)$,then $k + l =$

The perpendicular distance from the point $P(3, -2, 1)$ to the line joining the points $A(1, -3, 5)$ and $B(2, 1, -4)$ is:

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is $\qquad$

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

The foot of the perpendicular from the point $A(2, 0, 5)$ to the line $\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$ is $P(\alpha, \beta, \gamma)$. Then,which of the following is $NOT$ correct?

The point of intersection of the two lines $\frac{x - 3}{3} = \frac{y - 3}{-1}, z - 1 = 0$ and $\frac{x - 6}{2} = \frac{z - 1}{3}, y - 2 = 0$ is