The null points where electric field is zero must lie inside the triangle and on the axes of reflection symmetry.

Consider a triangle with side $2a$. Suppose a null point P lies on the axis through charge 3, as in the diagram below.

Because of symmetry, the horizontal components of the electric fields at P due to charges 1 & 2 cancel out. The vertical components of electric field for the 3 charges are

$E_1=\frac{K}{r_1^2}\sin\theta=\frac{Ky}{r_1^3}$

$E_2=E_1$

$E_3=\frac{K}{r_3^2}$

Using geometry we also have

$r_1^2=a^2+y^2$

$r_3=h-y$

$h=a\sqrt3$

For P to be a point where the total electric field is zero, we must have $E_1+E_2=E_3$. After substituting from the above equations and rearranging we get an expression containing the single variable $y$ and parameter $a$:

$2y(a\sqrt3-y)^2=(a^2+y^2)^{3/2}$

We wish to find the proportional distance of P from vertex 3, so let $h=1$. The equation becomes

$2y(1-y)^2=(\frac13+y^2)^{3/2}$

If you wish you can expand this into a polynomial equation of degree 6. It may be possible to obtain an analytic solution, but this will be extremely difficult. It is much easier to solve the equation numerically (eg using WolframAlpha ). The solutions are

$y\approx 0.143521$

$y\approx 3.58216$

$y=\frac13 = 0.333333...$

We must reject the 2nd solution which lies outside the triangle and is unphysical. The 3rd solution is the trivial one in which P lies at the centroid, ie $\frac13h$ from each side. The 1st solution is non-trivial : P lies approx $\frac17 h$ from the nearest side.

There are 3 non-trivial positions for P (one on each median) and 1 trivial position, making a total of 4 points at which the electric field is zero.