1. Let the common root be $\alpha$.

   Since $\alpha$ is a root of $x^2+2x+3=0,$ we have $\alpha^2+2\alpha+3=0.$

2. Also, $\alpha$ is a root of $ax^2+bx+c=0,$ so $a\alpha^2+b\alpha+c=0.$

3. Now, the polynomial $x^2+2x+3$ has discriminant $\Delta=2^2-4\cdot 1\cdot 3=4-12=-8<0.$

   Hence its roots are non-real and conjugate.

4. Since $a,b,c\in \mathbb{R}$, the polynomial $ax^2+bx+c$ also has real coefficients. If it has one non-real root in common with $x^2+2x+3$, then it must also have the conjugate root in common.

   Therefore, both roots of $x^2+2x+3$ are roots of $ax^2+bx+c$.

5. Since both are quadratic equations, the second polynomial must be a constant multiple of the first: $ax^2+bx+c=k(x^2+2x+3)$ for some real constant $k\neq 0$.

   Therefore, $a:b:c=1:2:3.$

6. Checking options:

   • A: $1:2:3$ ✅
   • B: $3:2:1$ ❌
   • C: $1:3:2$ ❌
   • D: $3:1:2$ ❌

So the correct answer is A.