Label the given equations as follows:
$$a + b + c = p \qquad (23)$$
$$a^2 + b^2 + c^2 = q \qquad (24)$$
$$bc - x - 2ya = 0 \qquad (25)$$
$$ca - x - 2yb = 0 \qquad (26)$$
$$ab - x - 2yc = 0. \qquad (27)$$

Adding $(25) \cdot (c - b)$, $(26) \cdot (a - c)$ and $(27) \cdot (b - a)$, we obtain
$$(a - b)(b - c)(c - a) = 0.$$

Hence one of $a - b, b - c, c - a$ must be zero. But if all three of $a, b, c$ are equal, to $a$, say, then $p = 3a, q = 3a^2$, so that $p^2 = 9a^2 = 3(3a^2) = 3q$ which has been ruled out. So, at most one of $a - b, b - c, c - a$ can be zero. Say $b = a$; then $c = p - 2a$ and
$$q = 2a^2 + c^2 = 2a^2 + (p - 2a)^2 = 6a^2 - 4ap + p^2,$$
so that $a$ satisfies the quadratic equation
$$6a^2 - 4pa + p^2 - q = 0, \quad \text{and so} \quad a = \frac{2p \pm \sqrt{2(3q - p^2)}}{6}.$$

We then have
$$c = p - \frac{2p \pm \sqrt{2(3q - p^2)}}{3} = \frac{p \mp \sqrt{2(3q - p^2)}}{3},$$

$$\begin{align*} abc &= a^2c = \frac{(2p \pm \sqrt{2(3q - p^2)})^2 (p \mp \sqrt{2(3q - p^2)})}{54} \\ &= \frac{5p^3 - 9pq \pm (p^2 - 3q)\sqrt{2(3q - p^2)}}{54} \end{align*}$$