Let $\triangle A_0B_0C_0$ be the antimedial triangle of $\triangle ABC.$ $P$ is the projection of $A_0$ on $B_0C_0,$ $M \equiv AG \cap BC$ is the midpoint of $BC,$ $U \equiv PA_0 \cap BC$ is the midpoint of $PA_0$ and $D \equiv GA' \cap B_0C_0$ is clearly the reflection of $G$ about $A'',$ since $\frac{_{\overline{GA'}}}{^{\overline{GD}}}=\frac{_{\overline{GM}}}{^{\overline{GA}}}=-\frac{_1}{^2}.$ Hence $A,A'',U$ are collinear, i.e. $A''$ lies on the A-cevian $AU$ of the isotomic conjugate $X_{69}$ of the orthocenter of $\triangle ABC.$ Likewise, $BB''$ and $CC''$ pass through $X_{69}.$ P.S. This property is valid for all the points lying on the Thomson cubic of ABC.

Let $D$ be the foot of the A-altitude.Let $AA''$ meets the the side $BC$ at $D'$.Now clearly $A'A''\parallel AD$.So clearly $D'G$ bisects $AD$.Let $A_1$ be the midpoint of $AD$.Note that $A_1, G, D'$ are collinear.Let $A_2$ be the reflection of $A$ wrt $G$.Then $A_1G\parallel A_2D\Longrightarrow A_2D\parallel D'G$.Now $BC$ bisects $G'A_2$.Hence $GDA_2D'$ is a parallelogram and so $D'$ is the isotomic point of $D$ wrt $BC$.Hence $AA'', BB'', CC''$ concur at the isotomic conjugate of the orthocenter of $ABC$ which is again the symmedian point of the anti-complementary triangle of $\triangle ABC$.

Here is my solution, though a bit long. Firstly, I will use a lemma. Lemma. If $A_1, B_1, C_1$ are the midpoints of $BC, CA, AB$ and if $A_2, B_2, C_2$ are the feet of perpendiculars from $G$ onto $BC, CA, AB$ such that $X,Y,Z$ are the midpoints of $GA_2, GB_2, GC_2;$ then the lines $A_1X, B_1Y, C_1Z$ concur at the symmedian point of $ABC.$ Proof. Let $AA_3$ be the $A$-altitude, and let $A_4$ be the midpoint of $AA_3.$ Then if $K$ is the symmedian point, it's well-known that $KA_1$ passes through $A_4.$ Homothety about $A_1$ that maps $AA_3$ to $GA_1$ maps the point $A_4$ to $X.$ So, $A_1A_4$ passes through $X.$ So, we are done.$\Box$ Coming back to our original problem, we see that a homothety about $G$ with ratio $-2$ maps $A_1B_1C_1$ to $ABC.$ This also maps the point $X$ to the point $A''.$So we finally get that, $AA'', BB'', CC''$ concur at the point $K'$ where $K'$ is the image obtained from homothety about $G$ with a ratio $-2$ of $K.$ We are done. $\Box$

Use barycentric coordinate. We get $A'=(0,3a^2+b^2-c^2,3a^2+c^2-b^2)$ and like wise. Reflection in G we get $A"=(4a^2,a^2+c^2-b^2,a^2+b^2-c^2)$ and like wise. So The equation of $AA"$ is $z(a^2+c^2-b^2)=y(a^2+b^2-c^2)$ and like wise. We can now easily use determinant method to show $AA",BB",CC"$ are concurrent. The point of concurrence is $(b^2+c^2-a^2,a^2+c^2-b^2,a^2+b^2-c^2)$ which is nothing but the isotomic conjugate of orthocentre.

(Sketch for 6/8/21 PoTD) Let $AA'' \cap BC = A_1$, the midpoint of $BC$ be $M$, the foot of the $A$-altitude be $D$, and the perpendicular from $A''$ to $AD$ have foot $X$. Using centroid length ratios, we find: $\bullet$ $\frac{A'G}{DA} = \frac{1}{3} \implies \frac{A'A''}{DA} = \frac{2}{3} \implies \frac{AA''}{AA_1} = \frac{1}{3};$ $\bullet$ $\frac{DA'}{DM} = \frac{AG}{AM} = \frac{2}{3};$ $\bullet$ $\frac{XA''}{DA_1} = \frac{DA'}{DA_1} = \frac{AA''}{AA_1} = \frac{1}{3}.$ Thus, $3 \cdot DA' = DA_1 = 2 \cdot DM \implies \frac{DM}{DA_1} = \frac{1}{2},$ or $AD$ and $AA_1$ are isogonal with respect to $ABC$. By symmetry, we can conclude $AA'', BB'', CC''$ concur at the isotomic conjugate of the orthocenter of $ABC$.