Solution:

The key claim is that we must have $\frac{1}{X G}+\frac{1}{Y G}+\frac{1}{Z G}=0$ (in directed lengths).

We present three proofs of this fact.

Proof 1: By a suitable affine transformation, we can assume without loss of generality that $A B C$ is equilateral. Now perform an inversion about $G$ with radius $G A=G B=G C$. Then the images of $X, Y, Z$ (call them $X', Y', Z'$) lie on $(G B C), (G A C), (G A B)$, so they are the feet of the perpendiculars from $A_1, B_1, C_1$ to line $X Y Z$, where $A_1, B_1, C_1$ are the respective antipodes of $G$ on $(G B C), (G A C), (G A B)$. But now $A_1 B_1 C_1$ is an equilateral triangle with medial triangle $A B C$, so its centroid is $G$. Now the centroid of (degenerate) triangle $X' Y' Z'$ is the foot of the perpendicular of the centroid of $A_1 B_1 C_1$ onto the line, so it is $G$. Thus $X' G + Y' G + Z' G = 0$, which yields the desired claim.

Proof 2: Let $W$ be the point on line $X Y Z$ such that $W G = 2 G X$ (in directed lengths). Now note that $(Y, Z ; G, W)$ is a harmonic bundle, since projecting it through $A$ onto $B C$ gives $(B, C ; M_{B C}, \infty_{B C})$. By harmonic bundle properties, this yields that $\frac{1}{Y G} + \frac{1}{Z G} = \frac{2}{W G}$ (in directed lengths), which gives the desired.

Proof 3: Let $P \neq G$ be an arbitrary point on the line $X Y Z$. Now, in directed lengths and signed areas, $\frac{G P}{G X} = \frac{[G B P]}{[G B X]} = \frac{[G C P]}{[G C X]}$, so $\frac{G P}{G X} = \frac{[G B P] - [G C P]}{[G B X] - [G C X]} = \frac{[G B P] - [G C P]}{[G B C]} = \frac{3([G B P] - [G C P])}{[A B C]}$. Writing analogous equations for $\frac{G P}{G Y}$ and $\frac{G P}{G Z}$ and summing yields $\frac{G P}{G X} + \frac{G P}{G Y} + \frac{G P}{G Z} = 0$, giving the desired.

With this lemma, we may now set $X G = g$ and know that
$$\frac{1}{g} + \frac{1}{g-5} + \frac{1}{g-8} = 0$$
Solving the quadratic gives the solutions $g = 2$ and $g = \frac{20}{3}$; the latter hence gives the maximum (it is not difficult to construct an example for which $X G$ is indeed $20/3$).