coplanar lines in 3d plane If we are given two lines 

Coplaner then how can we find the value of k.

I think if they are coplaner then their cross product should be zero . 



In the solution it is given as 

But I could not understand what they have done.
Hint: Examine both lines in parametric form. If their vectors are parallel then they are certainly coplanar. If their vectors are not parallel, two lines are coplanar if and only iff they intersect; otherwise, they are skew.  Hope it helps.
I think if they are coplaner then their cross product should be zero.


This statement is at least unclear, and not quite meaningful or correct. First of all, what do "they" and "their" refer to? If you mean the lines, then this statement is meaningless because cross product is an operation on vectors, not lines. It would be meaningful when talking about vectors, but then it's not true &mdash; the cross product of two (nonzero) vectors is zero means that they are collinear, not coplanar.

Let $\vec{a}$ and $\vec{b}$ be the directional vectors of two lines, and $A$ and $B$ some points on the same lines, respectively. (All of these can be seen from the given equations.) The condition given in the answers requires that the three vectors $\vec{a}$, $\vec{b}$, and $\vec{AB}$ be coplanar to guarantee that the two lines are. To check whether they are coplanar, we can set up the triple scalar product of these vectors. Effectively it finds the volume of the parallelepiped (up to the sign) built on these three vectors as sides. If the volume is zero, the parallelepiped is "flat" due to the three vectors lying in the same plane.

Why do we want $\vec{a}$, $\vec{b}$, and $\vec{AB}$ to be coplanar? Consider two cases. (1) If $\vec{a}\parallel\vec{b}$, then the lines are parallel and therefore coplanar. Note that in this case the three vectors are also coplanar regardless of the third vector. (2) Otherwise, we need to distinguish between intersecting lines (coplanar) and  skew lines (not coplanar). If the lines are intersecting, then all their points lie in the same plane as $\vec{a}$ and $\vec{b}$, therefore $\vec{AB}$ must lie in that same plane.