My question expands on a related question on the link, Why is the clique problem NP-complete?
In that post the author argued that while the $k$-clique problem is NP-complete; for a fixed $k$ the $k$-clique problem is polynomial time solvable. The question "Prove that the presence of 3-clique in a graph can be decided in polynomial time" also appears in the book,
S. K. Basu, "Design Method and Analysis of Algorithms, PHI, 2005, pg159.
To avoid any confusion, I felt it important to clarify that this would only be possible if the graph $G=(V,E)$ ($V$ is the set of vertices and $E$ is the set of edges) is fixed. In such a case a polynomial number of steps $R^m,$ (where $R=f(V,E)$, i.e. $R$ is a constant given by some function of vertices and edges) can be used to find the solution of 3-clique for example. So the conclusion that finding 3-clique for a fixed graph has polynomial time complexity works perfectly fine until this point.
But such a polynomial time algorithm with time complexity less than or equal to linear time complexity can not be designed to find 3-clique, i.e. $m>1$. Assuming that $k$ is fixed but $G$ is not fixed, in that case the complexity of finding 3-clique will then be exponential, unless someone can provide a polynomial time algorithm with complexity $R^m$, where $m\leq 1$ to find 3-clique for a fixed arbitrary graph. Since we are unable to prove that a polynomial time algorithm does not exist for 3-clique problem on an arbitrary graph, the computational complexity of the problem will hence be treated as being NP-complete based on the relationship $3SAT\leq_p k$-clique, a proof of which can be found in many standard books on algorithms.