Pain and gain

We can parameterize this circle as $x=a\cos(t)-b\sin(t),y=-a\cos(t)-b\sin(t),z=2b\sin(t)$ , where $a=\frac{1}{\sqrt{2}}$ and $b=\frac{1}{\sqrt{6}}$ . Now $xyz=\frac{2}{\sqrt{6}}\left(\frac{1}{6}\sin^2(t)-\frac{1}{2}\cos^2(t)\right)\sin(t)=-\frac{1}{3\sqrt{6}}\sin(3t)$ has a maximal absolute value of $\frac{1}{3\sqrt{6}}$ , and the maximum of $x^2y^2z^2$ is $\frac{1}{54}\approx 0.0185$ . The answer is $\boxed{185}$

Alternatively, for a given solution $x,y,z$ , consider the polynomial $f(t)=(t-x)(t-y)(t-z)$ $=t^3-\frac{1}{2}t-xyz$ , by Viete. Using a little calculus, we observe that the polynomial $t^3-\frac{1}{2}t-q$ has all real roots if and only if $|q|\leq\frac{1}{3\sqrt{6}}$ ; those values of $q$ are the possible values of $xyz$ .

Relevant wiki: Cubic Discriminant

Because $(x+y+z)^2 = x^2+y^2 + x^2 + 2(xy+xz+yz)$ , then $xy + xz + yz = - \dfrac12$ .

Let $f(X)$ be a monic cubic polynomial with roots $x,y$ and $z$ . Then by Vieta's formula , $f(X) = X^3 - \dfrac12 X + Q$ , where $Q = -xyz$ . So we want to maximize $Q^2 = P$ .

Since all it's roots are real numbers, then its cubic discriminant is non-negative:

$b^2 c^2 -4ac^3 - 4b^3 d - 27a^2 d^2 + 18abcd \geq 0 \; ,$

with $a= 0, b = 0, c = -\dfrac12, d = Q$ . Substituting these values and simplifying it gives $Q^2 \leq \dfrac1{54}$ . So $P = \dfrac1{54}$ , and so our answer is $\lfloor 10000 P \rfloor = \left \lfloor \dfrac{10000}{54} \right \rfloor = \boxed{185}$ .