Try the general case first

The most convenient point for calculating the actual length of a side of the triangle is a vertex. The sum of squares of distances translates in this case into $2a^2$ where $a$ is the length of the side.

$2a^2=98$ is satisfied by $a=\boxed{7}$

What's left is to show that the sum of the squares of the distances is independent of the point chosen.

For this it is convenient to pick an equilateral triangle with vertices on a unit circle, namely points $(0, 1), (\frac{\sqrt{3}}{2}, -\frac{1}{2}), (-\frac{\sqrt{3}}{2}, -\frac{1}{2})$

A general point $(x,y)$ on the unit circle will satisfy the equation $x^2+y^2=1$

The sum of the squares of distance of this point from the vertices will be

$(x-0)^2+(y-1)^2+(x-\frac{\sqrt{3}}{2})^2+(y+\frac{1}{2})^2+(x+\frac{\sqrt{3}}{2})^2+(y+\frac{1}{2})^2=$

$y^2-2y+1+2y^2+2y+\frac{2}{4}+x^2+2x^2+2\times\frac{3}{4}=$

$3x^2+\frac{3}{2}+3y^2+\frac{3}{2}=$

$3(x^2+y^2)+3=6$

By replacing $x^2+y^2$ with $1$ , the last dependence on the location has been removed.

For a triangle $\triangle ABC$ , with centroid $G$ and any given point P, we have the following relation:

$PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^2 + 3PG^2$ So, in case of an equilateral triangle, this reduces to $PA^2 + PB^2 + PC^2 = 6R^2 = 2l^2$ , where $R$ is the radius of the circumcircle and $l$ is the side length of the equilateral triangle.