$a^2-bc-(b^2-ca)=(a-b)(a+b+c) \to a=b$ or $a+b+c=0$ If $a+b+c \neq 0$ then $a=b=c \to a^2-bc =0 \to $ contradiction If $a+b+c=0$ then $6=a^2+b^2+c^2-(ab+bc+ca)=a^2+b^2+c^2-(ab+bc+ca)-(a+b+c)^2=-3(ab+bc+ca) \to ab+bc+ca=-2$ And we can choose $(a,b,c)=(0, \sqrt{2},-\sqrt{2})$ PS. $bc=a^2-2 \leq \frac{(b+c)^2}{4}=\frac{a^2}{4} \to 3a^2-8 \leq 0$ So let $a$ is any real such that $|a| \leq \frac{2\sqrt{6}}{3}$ and $b,c$ are roots of $x^2+ax+a^2-2=0 \to b=\frac{-a +\sqrt{8-3a^2}}{2},c=\frac{-a-\sqrt{8-3a^2}}{2}$

$a^2-bc=2 \to a^3-abc=2a$ $b^2-ca=2 \to b^3-abc=2b$ $c^2-ab=2 \to c^3-abc=2c$ , and keep in mind that $(1) a^2+b^2+c^2-ab-bc-ca=6$ Sum these three equations to get $a^3+b^3+c^3-3abc=2(a+b+c), a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ We have two cases; $a+b+c$ is nonzero $\implies a^2+b^2+c^2-ab-bc-ca=2$,which is a contradiction to $(1)$.So $a+b+c=0$. $(a+b+c)^2=0 \to a^2+b^2+c^2=-2(ab+ac+bc)$ and replace that in $(1)$ to get $ab+ca+bc=-2$. One triplet that would satisfy these conditions would be $(a,b,c)=(0, \sqrt{2},-\sqrt{2})$

2021 JBMO TST Bosnia and Herzegovina P1: Determine all real numbers $a, b, c, d$ for which $$ab + c + d = 3$$$$bc + d + a = 5$$$$cd + a + b = 2$$$$da + b + c = 6$$