If I place 3 equal charges on an equilateral triangle, I get 4 equilibrium positions. How do i draw an electric field diagram representing this? 
*

*The equilibrium at the centre


We know that there is an equilibrium position at the centre,by symmetry. We can also verify it using vector addition of forces. 


*

*The 3 non-standard equilibrium positions 


The other 3 positions can be figured out qualitatively by the opposite direction of forces at the centre,when displaced in the direction of a vertice ,and when the test charge is kept at the midpoint of the side opposite to the charge. By symmetry, we will have 3 such positions. 


*

*Drawing the diagram 


However, when I try to draw a field diagram, I cannot seem to find a suitable arrangement in which the lines can be drawn, without intersecting each other, that would accommodate for the 3 non standard unstable equilibriums. Kindly Help.
 A: To check your reasoning, I made a plot of the potential along the line from a vertex through the centre

confirming what you say.
This being so, I sketched, for the field lines

A surface plot of the potential is

The field lines are lines of greatest slope
A: 
However, when I try to draw a field diagram, I cannot seem to find a suitable arrangement in which the lines can be drawn, without intersecting each other.

That is correct. The field lines intersect each other at all of the equilibrium points: 


*

*At the center of the triangle, there is a neighbourhood (which includes a disk of radius $\sim 0.2$ if the triangle has side $\sqrt{3}$) in which every field line terminates at the center. There are also two field lines going out of the plane of the triangle along the normal of that plane.

*At the other equilibrium points, inside the plane of the triangle, you have two incoming and two outgoing field lines which meet at the equilibrium point. (If you consider the out-of-plane behaviour, however, there is an infinity of outgoing field lines which initiate at these equilibrium points.)


The reason we say that field lines cannot cross is that they are defined as curves $s\mapsto \vec\gamma(s)$ which are tangent to the electric field, i.e.
$$
\frac{\mathrm d}{\mathrm ds}\vec\gamma(s) = \vec E(\vec\gamma(s)),
\tag 1
$$
and therefore


*

*if two field lines pass through the same point $\vec r_0$ then they have the same tangent at that point, so

*they are solutions of the same first-order ODE $(1)$ 

*with the same initial condition,

*so they must be identical.


However, this argument fails at equilibrium points where $\vec E(\vec r_0) = \vec 0$. At those points, the generic behaviour is that field lines will point into (or out of) the equilibrium point, at which point they terminate:


*

*If you define the field lines as in $(1)$, then $\vec\gamma(s)$ slows down without limit, and $\vec\gamma(s)\to \vec r_0$ as $s\to \infty$, and the field line covers an open interval that approaches $\vec r_0$ arbitrarily closely.

*If instead you choose to re-parametrize the field lines by path length, adapting $(1)$ to 
$$
\frac{\mathrm d}{\mathrm ds}\vec\gamma(s) = \frac{1}{|\vec E(\vec\gamma(s))|}\vec E(\vec\gamma(s)),
\tag 2
$$
then the right-hand side has a discontinuity at $\vec\gamma(s)=\vec r_0$, and the ODE cannot be continued past that point.

*If you actually try to initiate a field line as per $(1)$ at the equilibrium point where $\vec E(\vec r_0)=\vec 0$, then $\vec\gamma(s)\equiv \vec r_0$ will just stay there forever.
In the neighbourhood of an equilibrium point, most field lines will miss the equilibrium point, and they will look like hyperbolas when they are close to it. However, there will generically be a small set (in technical language, a set 'of measure zero') of lines that touch the equilibrium point, tending to it as $s$ increases or decreases.
As to what the field-line diagram actually looks like, the sketch provided in the answer by Charles Francis is an accurate rendition (though it's missing the crossings at the off-center equilibrium points).
