Label the given equations as follows:
$$a + b + c = 1 \tag{10}$$
$$a^2 + b^2 + c^2 = 3 \tag{11}$$
$$bc - x - 2ya = 0 \tag{12}$$
$$ca - x - 2yb = 0 \tag{13}$$
$$ab - x - 2yc = 0. \tag{14}$$

Solution 1. Eliminate $x$ by subtracting (13) from (12), and (14) from (13) to get $c(b-a) = 2y(a-b)$ and $a(c-b) = 2y(b-c)$. From these it follows that
$$c(b-a)(b-c) = 2y(a-b)(b-c) = a(c-b)(a-b),$$
i.e.,
$$0 = a(c-b)(a-b) - c(b-a)(b-c) = (a-b)(b-c)(c-a).$$

So, one of $a-b, b-c, c-a$ must be zero. Wlog, suppose that $a=b$. Then, by (10) and (11),
$$2a + c = 1 \quad \text{and} \quad 2a^2 + c^2 = 3.$$
Hence $3 = 2a^2 + (1-2a)^2 = 6a^2 - 4a + 1$ and so $0 = (a-1)(3a+1)$. If $a=1$, we have $a=b=1$ and $c = 1 - 2a = -1$ which gives $abc = -1$. If $a = -1/3$, we have $a = b = -1/3$, $c = 1 - 2a = 5/3$ and so $abc = 5/27$.

Solution 2. From (10) and (11) we obtain
$$-2 = (a + b + c)^2 - (a^2 + b^2 + c^2) = 2(ab + bc + ca), \text{ i.e.}$$
$$ab + bc + ca = -1. \tag{15}$$

Adding (12), (13) and (14) together, and using (10) and (15) we get
$$3x + 2y = -1. \tag{16}$$

Multiplying (12), (13) and (14) by $a, b, c$, respectively, then adding these together and using (10) and (11) gives us
$$x + 6y = 3abc. \tag{17}$$

$$\begin{align*} 1 &= (ab + bc + ca)^2 && \text{using (15)} \\ &= (ab)^2 + (bc)^2 + (ca)^2 + 2abc(a + b + c) \\ &= (x + 2ya)^2 + (x + 2yb)^2 + (x + 2yc)^2 + 2abc && \text{by (12), (13), (14)} \\ &= 3x^2 + 4xy + 12y^2 + 2abc && \text{using (10), (11)} \\ &= x(3x + 2y) + 2y(x + 6y) + 2abc \\ &= -x + 6yabc + 2abc && \text{using (16), (17)} \\ &= 6y - 3abc + 6yabc + 2abc && \text{using (17) again} \\ &= 6y(abc + 1) - abc. \end{align*}$$

$$0 = 6y(abc + 1) - abc - 1 = (6y - 1)(abc + 1).$$
Therefore, either $abc = -1$, or $y = 1/6$. In the latter case, equation (16)
gives $x = -4/9$ and then (17) implies $abc = 5/27$.