condition for the existence of lines on degree four hypersurface in P^3.. i just wanted to know some necessary and sufficient conditions
 A: The point of this answer is to show that the answer should be a degree 320 hypersurface in $\mathbb{P}^{34}$. This suggests that you should NOT want to compute the answer explicitly if you can possibly avoid it. On the other hand, as I suggest in the comments to my other answer, it should be tractable to check whether individual points are in it. 
Note: This is a standard example of how to do computations with Chern classes. I'm not going to explain all the terminology I use. The standard references are Griffiths and Harris, or Fulton. There was also a superb course taught by Joe Harris at Harvard in 2002. There used to be notes from it online; I don't know if they are still available somewhere.
So: Let $G(2,4)$ be the Grassmannian of lines in $\mathbb{P}^3$. Let $X \to G(2,4) \times \mathbb{P}^3$ be the incidence variety of pairs (line, point on line) in $\mathbb{P}^3$. So $X \to G(2,4)$ is a $\mathbb{P}^1$-bundle.
Then we can pull back $\mathcal{O}(4)$ from $\mathbb{P}^3$ to $X$, getting a line bundle, and we can push this line bundle forward from $X$ to $G(2,4)$ getting a rank $5$ vector bundle $V$ on $G(2,4)$. Conceptually, for $\ell$ a line in $\mathbb{P}^3$, the fiber of $V$ above the point $[\ell]$ of $G(2,4)$ is $H^0(\ell, \mathcal{O}(4)|_{\ell})$. 
In particular, if $F$ is any section in $H^0(\mathbb{P}^3, \mathcal{O}(4)$, we can pull-push $F$ to a section of $V$ on $G(2,4)$. This section, on $\ell$, is $F|_{\ell}$ and, in particular, vanishes at $[\ell]$ if and only if $F$ vanishes on $\ell$.
Now, a generic $F$ will not vanish on any line. We want to know the degree of the hypersurface of $F$'s that will. In other words, given two sections $F_1$ and $F_2$ of $V$, we want to know how many of the linear combinations $a_1 F_1 + a_2 F_2$ have zeroes somewhere on $G(2,4)$. Now, such $a_1$ and $a_2$ exist if and only if $F_1$ and $F_2$ are linearly dependent somewhere on $G(2,4)$. What one expects (but needs to prove) is that, for each point of $G(2,4)$ where $F_1$ and $F_2$ become linearly dependent, one finds exactly one degree $4$ form $a_1 F_1  + a_2 F_2$ (up to scalar) such that $a_1 F_1 + a_2 F_2$ vanishes at that point. 
Now, the places where $2$ sections of $V$ become linearly dependent are counted (with multiplicity) by the Chern class $c_4(V)$. One expects that there are no multiplicites, so one wants to compute $c_4(V)$. 
The cohomology ring of $G(2,4)$ is 
$$\mathbb{Z}[u, v]^{\mathrm{sym}}/(u^3+u^2v+uv^2+v^3,\ u^4+u^3v+u^2v^2+u v^3+ v^4)$$
where the superscript means to only take symmetric polynomials in $u$ and $v$.
In particular, $H^4(G(2,4))$ is spanned by $u^4+v^4$, $u^3 v + u v^3$, $u^2 v^2$ modulo the relations $u^3 v + u v^3 = - u^2 v^2$ (because  $u^3 v + u^2 v^2 + u v^3 = (u+v)(u^3+u^2 v + u v^2 + v^3) - (u^4+u^3v+u^2v^2+u v^3+ v^4)$) and $u^4 + v^4 = 0$ (because $u^4 + v^4 = -(u+v)(u^3+u^2 v + u v^2 + v^3) +2 (u^4+u^3v+u^2v^2+u v^3+ v^4)$).
The Chern roots of $V$ are $4u$, $3u+v$, $2u+2v$, $u+3v$ and $4v$. (Write $V = \mathrm{Sym}^4 (S^*)$, where $S$ is the tautological bundle, and recall that the Chern roots of $S$ are $-u$ and $-v$.)  So $c_4(V)$ is the fourth degree term in $(1+4u)(1+3u+v)(1+2u+2v)(1+u+3v)(1+4v)$. Thanks to Mathematica, this is $24 u^4 + 304 u^3 v + 624 u^2 v^2 + 304 u v^3 + 24 v^4$. Modulo the relations $u^3 v + u v^3 = - u^2 v^2$ and $u^4 + v^4 = 0$, this is $320 u^2 v^2$. Recalling that $u^2 v^2$ is the fundamental class, we get that there are $320$ lines on which $F_1$ and $F_2$ become linearly dependent, so the hypersurface in question has degree $320$.
A: I don't know if there is a necessary and sufficient condition, but there are a few things you can say. Let's say $X$ is a $K3$
1) If a $X$ it contains a line, then it contains a smooth rational curve
2) If a $X$ contains a smooth rational curve $C$, then $C^2=-2$.
3) If $X$ contains a smooth rational curve, then its Picard number is at least $2$.

4) If there is any effective $1$-cycle on $X$ whose self-intersection is negative, then $X$ contains a class with self-intersection $-2$.
5) If $X$ contains a class with self-intersection $-2$, then it contains a smooth rational curve.
6) If the Picard number of $X$ is at least $12$, then it contains a smooth rational curve.

7) A degree four surface in $\mathbb P^3$ is a $K3$ with a very ample Cartier divisor $H$ such that $H^2=4$ and $\dim |H|=4$.
8) A $K3$ surface embedded into $\mathbb P^3$ as a degree four surface contains a line if and only if it contains an effective $1$-cycle $L$ with $L^2=-2$ and $H\cdot L=1$ where $H$ is the very ample line bundle from 5). (Edit: originally here I said "indecomposable" $1$-cycle, but I just realized that the condition $H\cdot L=1$ implies that) Unfortunately, I don't think this is a very useful criterion as this is just a reformulation of the fact that $L$ is a line on a degree four surface in $\mathbb P^3$.

Summary there are some relatively good criteria to check whether or not $X$ contains a smooth rational curve, at least if you think that knowing the Picard number, or the range of self-intersection numbers is a good criterion. It is harder to tell whether it is a line. 
On the other hand, if you make up any even quadratic form of signature $(1,\varrho-1)$ on a lattice of rank at most $11$, then it will occur as the Picard lattice of a $K3$, so if you ensure that your quadratic form has an $H$ and an $L$ as above, then there is a good chance that there is a degree four surface with that Picard lattice. It is not entirely certain of course, because the lattice does not tell you $\dim|H|$ but I think it suggests that there are many possibilities. 
Another interesting fact is that if you take any degree four surface that contains two lines contains many other smooth rational curves as well. In other words, knowing that locating smooth rational curves is very far from locating lines.
A: Here's how to obtain such a condition in principle. Doing it in practice will require a computer algebra system, and may need a lot of computer power:
The generic line in $\mathbb{P}^3$ can be parameterized as $(x:y:ax+by:cx+dy)$ for some $(a,b,c,d)$. More specifically, this is the dense Schubert cell in $G(2,4)$. Let your degree $4$ polynomial be $F$. Write
$$F(x,y,ax+by,cx+dy) = A(a,b,c,d) x^4 + B(a,b,c,d) x^3 y + \cdots + E(a,b,c,d) y^4.$$
So $A$, $B$, $C$, $D$ and $E$ are $5$ polynomials in $4$ variables, each non-homogenous with maximal degree $4$. Let their resultant be $R$. (If you haven't heard of multivariate resultants, skip to the end.) Then $R=0$ is a necessary condition for $F$ to contain a line. There will most likely be some factor $Q$ of $R$ which is a necessary and sufficient condition, but I don't immediately see a better way to obtain $Q$ than by factoring $R$ and trying points on each irreducible component.
Why does this happen? Well, if there is a line $\ell$ in $F=0$, and $\ell$ lies in the dense cell of $G(2,4)$, then $(A,B,C,D,E)$ have a common root so $R$ vanishes. If we have a pair $(F, \ell)$, but $\ell$ is not in the dense cell, then let $g(t)$ be a generic path in $PGL_4$, with $g(0) = \mathrm{Id}$. Then $g(t) \ell$ is in the dense cell for $t \neq 0$, and the preceeding argument shows that $R$ vanishes on $g(t)^* F$. By continuity, $R$ vanishes on $F$.
Why isn't this argument invertible? The vanishing of $R$ is equivalent to the homogenized versions of $(A,B,C,D,E)$ having a common root in $\mathbb{P}^4$. But what we want to know is whether the hypersurfaces they define have a common root in $G(2,4)$. There is a birational map $G(2,4)$ to $\mathbb{P}^4$, but it is not an isomorphism, so the hypersurfaces might meet in $\mathbb{P}^4$ but become disjoint in $G(2,4)$. I don't know for sure whether this happens in this case, but it often does. The good thing is that we suspect that the actual locus we are hunting for is codimension $1$, so we can find it as a factor of the polynomial $R$.
There is a computation we should do before this one, because it will help check for errors and tell us what to expect. We should compute what we expect the degrees of $Q$ and $R$ to be. EDIT: In my pother answer, I show that $Q$ should have degree $320$.
Finally, I wanted to point you to a quick online reference for computing resultants of multi-variate polynomials, but all the sources I checked only had the one variable case (or, equivalently, two variables with homogenous polynomials). The famous reference here is the book of Gelfand, Kapranov and Zelevinsky, but that is quite an undertaking. The one thing I will do is alert you to the Cayley trick: The resultant of $(A,B,C,D,E)$ is the same as the discriminant of 
$$A(a,b,c,d) + B(a,b,c,d) w + \cdots + E(a,b,c,d) z$$
where $(w,x,y,z)$ are auxilliary variables. So you can compute a disciminant in $8$ variables instead of a resultant in $4$, if your software prefers that.
