Note that $f(x)=x^2-(a+b+c)x+(ab+bc+ca)=0$ is the required polynomial. Hence $f(a+b+c)=ab+bc+ca$.

$xf(x)-abc$ has three roots a,b,c so $xf(x)-abc=k(x-a)(x-b)(x-c)\Rightarrow (a+b+c)f(a+b+c)=k(a+b)(b+c)(c+a)+abc...$ note that $f(x)=\frac{k(x-a)(x-b)(x-c)+abc}{x}$ wont be a polynomial unless $k=1$, so the desired answer is $f(a+b+c)=ab+bc+ca$.

Let $f(x)=px^2+qx+r,$ then $$pa^2+qa+r=bc \;\;\; (1)$$$$pb^2+qb+r=ca \;\;\; (2)$$$$pc^2+qc+r=ab \;\;\; (3)$$Solving this system of equation yields us $$p=1$$$$q=-(a+b+c)$$$$r=bc+ca+ab $$ Hence, the required polynomial is $f(x)=x^2-(a+b+c)x + (bc+ca+ab).$ $$\boxed{\therefore f(a+b+c)=bc+ca+ab}.$$

The cubic polynomial $xf(x) - abc$ has roots $a,b,c$ by the given conditions. Thus it is monic since constant term = -product of roots = $-abc$. Thus $xf(x) - abc = (x-a)(x-b)(x-c) \implies f(x) = x^2 - (a+b+c)x + (ab+bc+ca) \implies f(a+b+c) = \boxed{ab+bc+ca}$

Consider W.L.O.G $a=0$ then we have $f(b)=f(c)=0$ , and $f(0)=bc$ then clearly $f(x)=(x-b)(x-c)$ so we have $f(a+b+c)=(a+c)(a+b)=a^2+(b+c)a+bc=bc \equiv ab+bc+ca$ Now we consider none of $a,b,c$ is $0$ , so we have $xf(x)-abc=k(x-a)(x-b)(x-c)$ , where $k \in \mathbb{R} \setminus \{0\}$ , now in that case we have $-abc=k(-abc) \implies k=1$ , now clearly $P(x)=xf(x)-abc$ has no root as $0$ because if $0$ is a root then $P(0)=-abc=0 \implies $one of $a,b,c$ is $0$ which is not possible , hence we have $f(x)=x^2-(a+b+c)x+ab+bc+ca$ , so $f(a+b+c)=ab+bc+ca$. $\blacksquare$

Just bash by Lagrange interpolation

Easy problem! Let $f(x)=px^2+qx+r$. The three given equations become, $pa^2+qa+r=bc$....$(1)$ $pb^2+qb+r=ac$....$(2)$ $pc^2+qc+r=ab$....$(3)$ Subtract $(2)$ from $(1)$, we get: $p(b+a)+q+c=0$. By symmetry, we have another equation $p(b+c)+q+a=0$. Equating the LHS of the two equations, we get that $p=1$. Also, we have that $q=-(a+b+c)$. Substituting the value of $p$ an $q$ in $(1)$, we get that $r=ab+bc+ca$. So, $f(x)=x^2-(a+b+c)x+ab+bc+ca$. Trivially, we have that $f(a+b+c)=ab+bc+ca$

f(x) be a polynomial with real coefficients of degree 2 such that f(a) = bc,f(b) = ca and f(c) = ab Notice that, a.f(a) = b.f(b) = c.f(c) = abc Now, x.f(x) = k(x - a)(x - b)(x - c) + abc where k is the leading coefficient. On plugging x = 0 in the above equation, we get k = 1 So, x.f(x) = (x - a)(x - b)(x - c) + abc x.f(x) = (x² - ax - bx + ab)(x - c) + abc x.f(x) = (x³ - ax² - bx² + abx - cx² + acx + bcx - abc) + abc x.f(x) = x³ - (a + b + c)x² + (ab + bc + ca)x f(x) = x² -(a + b + c)x + ab + bc + ca Now, On plugging x = a + b + c in the above equation f(a + b + c) = (a + b + c)² -(a + b + c)(a + b + c) + ab + bc + ca Therefore, f(a + b + c) = ab + bc + ca and we're done