how to find three vertices of a triangle.

Where s=circumcenter, H= orthocenter, and A'= midpoint of one side of triangle.

How can can I determine the location of the three vertices of the triangle?

Let us impose coordinates to facilitate the discussion. We will place $H$ at the origin which then gives $A' = (-1,\ 6)$ and $S=(-6,6)$.

From the geometry of the Euler line, we know that the centroid of the triangle $G$ lies on the line $SH$ in the ratio $|SG|:|GH| = 1:2$. Therefore the coordinate of $G$ is given by $G = (4,4)$.

We also know that $A'$ is the midpoint of a side, call it $a$. Since $S$ is the circumcenter of the triangle, it follows that the line $\overline{SA'}$ must be the perpendicular bisector of side $a$. This means that $a$ is perpendicular to $\overline{SA'}$ and in particular $a$ is parallel to the $y$-axis. Then the line perpendicular to $a$ and through $H$ must be an altitude of the triangle, i.e. the $x$-axis is an altitude of the triangle. Therefore the vertex $A$ opposite side $a$ must lie on the $x$-axis.

But $A$ must also lie on the line $\overline{A'G}$ for the centroid lies on lines joining vertices and midpoints of opposite sides. It follows that $A$ is the intersection of $\overline{A'G}$ with the $x$-axis and in particular we have $A = (-14,\ 0)$ in our case.

Finally, the circumcircle $\Gamma$ is the circle with center $S$ passing through $A$. The intersection of $\Gamma$ with the side $a$ then determines the remaining two vertices of the triangle. This determines the triangle uniquely given $A',\ S$ and $H$ and we can verify that each of the points are indeed the corresponding special points of the triangle.

Another way of solving the problem is to use vector algebra.

Let the coordinate system have origin at the circumcenter $\mathbf{S}=(x_s,y_s)=(0,0)$. The orthocenter has position vector $\mathbf{H}=(x_h,y_h)=(6,-6)$. Let $\mathbf{X}_1=(x_1,y_1), \mathbf{X}_2=(x_2,y_2),\mathbf{X}_3=(x_3,y_3)$ be the three vertices and let $\mathbf{A}=(x_a,y_a)=(7,0)$ be the the midpoint of line $23$.

Then $\mathbf{A} = 0.5 (\mathbf{X}_2+\mathbf{X}_3)$ and we can express $x_2$ and $y_2$ as $$x_2 = 14 - x_3 ~,~~ y_2 = -y3 \,.$$ Also, using the definition of the orthocenter and the vector scalar product, $$(\mathbf{X}_3-\mathbf{H})\cdot(\mathbf{X}_1-\mathbf{X}_2) = 0 ~,~~ (\mathbf{X}_2-\mathbf{H})\cdot(\mathbf{X}_1-\mathbf{X}_3) = 0 \,.$$ We can solve these two equations to express $(x_1,y_1)$ in terms of $(x_3,y_3)$ after eliminating $(x_2,y_2)$ to get \begin{align} x_1 &= \frac{y_3^3 + [x_3(x_3-14)+6]\,y_3 + 26(x_3-7)}{6x_3-y_3-42} \\ y_1 &= \frac{(7-x_3)y_3^2 + 6y_3- x_3^3 + 21 x_3^2 - 146x_3 + 336}{6x_3-y_3-42} \end{align} To solve for $x_3$ we can use the relation $$(\mathbf{S}-\mathbf{A})\cdot(\mathbf{X}_3-\mathbf{X}_2) = 0 \,.$$ The solution is $x_3 = 7$. To find $y_3$ we can use the definition of the circumcenter to get $$(\mathbf{X}_1-\mathbf{S})\cdot(\mathbf{X}_1-\mathbf{S}) = (\mathbf{X}_3-\mathbf{S})\cdot(\mathbf{X}_3-\mathbf{S})$$ The result is the equation $$y_3^4 - 87 y_3^2 + 1836 = 0$$ with solutions $(-7.14,-6,6,7.14)$. If we assume that $y_3 >0$ we have two possible values only one of which, $y_3 = 7.14$, is a valid solution. Using this solution we get $(x_1,y_1)=(-8,-6)$, $(x_2,y_2)=(7,-7.14)$, $(x_3,y_3)=(7,7.14)$.