Step 1: State the centroid formula in 3D.
For triangle $ABC$ with vertices $A(x_1,y_1,z_1)$, $B(x_2,y_2,z_2)$, $C(x_3,y_3,z_3)$, the centroid is: \[G = \left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3},\ \frac{z_1+z_2+z_3}{3}\right)\]
Step 2: Set up the coordinate equations.
With $A(4,x,1)$, $B(y,-5,2)$, $C(7,8,3)$, $G(3,5,2)$: \[\frac{4+y+7}{3} = 3, \quad \frac{x-5+8}{3} = 5, \quad \frac{1+2+3}{3} = 2\] The third equation is automatically satisfied ($6/3 = 2$).
Step 3: Solve for $y$ (the x-coordinate of B).
\[\frac{11+y}{3} = 3 \implies 11+y = 9 \implies y = -2\] So $B = (-2, -5, 2)$.
Step 4: Solve for $x$ (the y-coordinate of A).
\[\frac{x+3}{3} = 5 \implies x+3 = 15 \implies x = 12\] So $A = (4, 12, 1)$.
Step 5: Understand why $F$ is the midpoint of $AB$.
In any triangle, the centroid $G$ lies on each median and divides it in the ratio $2:1$ from vertex to opposite midpoint. The median from $C$ goes to the midpoint $F$ of $AB$. Since $CG$ extended meets $AB$ at $F$, $F$ must be the midpoint of $AB$.
Step 6: Find the midpoint $F$ of $AB$.
$A = (4,12,1)$ and $B = (-2,-5,2)$: \[F = \left(\frac{4+(-2)}{2},\ \frac{12+(-5)}{2},\ \frac{1+2}{2}\right) = \left(1,\ \frac{7}{2},\ \frac{3}{2}\right)\]
Step 7: State the final answer.
\[ \boxed{F = \left(1,\ \frac{7}{2},\ \frac{3}{2}\right)} \]