Condition for Two Lines to be on the Same Plane

Q. A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is $\frac{1}{4}$. Three stones $A,B\text{and}C$ are placed at the points $(1,1),(2,2)\text{and}(4,4)$ respectively. Then which of these stones is/are on the path of the man?

1. $C$only
2. the three
3. $A$only
4. $B$only

Q. The lines $\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta }$ and $\frac{x-b+c}{\beta -\gamma }=\frac{y-b}{\beta }=\frac{z-b-c}{\beta +\gamma }$ are coplanar and then equation to the plane in which they lie, is

1. $x+y+z=0$
2. $x-y+z=0$
3. $x-2y+z=0$
4. $x+y-2z=0$

Q. If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{1}=\frac{z}{1}$ intersect, then k=

1. $\frac{2}{9}$
2. $\frac{9}{2}$
3. $0$
4. None of these

Q.

Two lines $L_1:x=5,\frac{y}{3-\alpha }=\frac{z}{-2}$ and $L_2:x=\alpha ,\frac{y}{-1}=\frac{z}{2-\alpha }$ are coplanar. Then, $\alpha$ can take value(s)

1. 1

2. 2

3. 3

4. 4