There exist two complex numbers $c$, say $c_1$ and $c_2$, so that $-5 + 3i$, $8 - i$, and $c$ form the vertices of an equilateral triangle.  Find the product $c_1 c_2$.

[asy]
unitsize(0.5 cm);

pair A, B;
pair[] C;

A = (2,2);
B = (5,1);
C[1] = rotate(60,A)*(B);
C[2] = rotate(60,B)*(A);

draw(A--C[1]--B--C[2]--cycle);
draw(A--B);

dot("$-5 + 3i$", A, W);
dot("$8 - i$", B, E);
dot("$c_1$", C[1], N);
dot("$c_2$", C[2], S);
[/asy]
Let $a = 2 + 2i$ and $b = 5 + i$.  Let $\omega = e^{i \pi/3}$.  Then $\omega^3 = e^{i \pi} = -1$, so $\omega^3 + 1 = 0$, which factors as
\[(\omega + 1)(\omega^2 - \omega + 1) = 0.\]Since $\omega \neq -1$, we have that $\omega^2 - \omega + 1 = 0$.

We can obtain the complex number $c_1$ by rotating the number $b$ around the number $a$ counter-clockwise by $\pi/3$.

[asy]
size(100);

pair A, B;
pair[] C;

A = (2,2);
B = (5,1);
C[1] = rotate(60,A)*(B);
C[2] = rotate(60,B)*(A);

draw(B--A--C[1]);
draw(interp(A,B,0.3)..interp(A,rotate(30,A)*(B),0.3)..interp(A,C[1],0.3),Arrow(8));

dot("$a$", A, W);
dot("$b$", B, E);
dot("$c_1$", C[1], N);
label("$\frac{\pi}{3}$", interp(A,rotate(30,A)*(B),0.3), E);
[/asy]

This gives us the equation
\[c_1 - a = \omega (b - a),\]so $c_1 = \omega (b - a) + a$.

Similarly, we can obtain the complex number $c_2$ by rotating the number $a$ around the number $b$ counter-clockwise by $\pi/3$.

[asy]
size(100);

pair A, B;
pair[] C;

A = (2,2);
B = (5,1);
C[1] = rotate(60,A)*(B);
C[2] = rotate(60,B)*(A);

draw(A--B--C[2]);
draw(interp(B,A,0.3)..interp(B,rotate(30,B)*(A),0.3)..interp(B,C[2],0.3),Arrow(8));

dot("$a$", A, W);
dot("$b$", B, E);
dot("$c_2$", C[2], S);
label("$\frac{\pi}{3}$", interp(B,rotate(30,B)*(A),0.3), W);
[/asy]

This gives us the equation
\[c_2 - b = \omega (a - b),\]so $c_2 = \omega (a - b) + b$.

Then
\begin{align*}
c_1 c_2 &= [\omega (b - a) + a][\omega (a - b) + b] \\
&= -\omega^2 (a - b)^2 + \omega a(a - b) + \omega b(b - a) + ab \\
&= -\omega^2 (a - b)^2 + \omega (a - b)^2 + ab.
\end{align*}Since $\omega^2 - \omega + 1 = 0$ ($\omega$ is a primitive sixth root of unity), we have $\omega^2 = \omega - 1$, so
\begin{align*}
c_1 c_2 &= (1 - \omega) (a - b)^2 + \omega (a - b)^2 + ab \\
&= (a - b)^2 + ab \\
&= a^2 - ab + b^2.
\end{align*}Substituting $a = -5 + 3i$ and $b = 8 - i$, we get
\[c_1 c_2 = (-5 + 3i)^2 - (-5 + 3i)(8 - i) + (8 - i)^2 = \boxed{116 - 75i}.\]