If the equations $$x^2 + 2x + 3 = 0$$ and $$ax^2 + bx + c = 0$$, $$a, b, c \in R$$, have a common root, then $$a : b : c$$ is:

Let the two quadratic equations be

$$x^2 + 2x + 3 = 0 \qquad\text{and}\qquad ax^2 + bx + c = 0,$$

where $$a,\,b,\,c\in\mathbb R.$$

Suppose the number $$\alpha$$ is a common root of both equations. By definition of a root we have

$$\alpha^2 + 2\alpha + 3 = 0 \qquad\text{and}\qquad a\alpha^2 + b\alpha + c = 0.$$

From the first equation we can express $$\alpha^2$$ in terms of $$\alpha$$:

$$\alpha^2 = -2\alpha - 3.$$

Now we substitute this expression for $$\alpha^2$$ into the second equation. Doing so gives

$$a(-2\alpha - 3) + b\alpha + c = 0.$$

Expanding the product and collecting like terms in $$\alpha$$ we obtain

$$(-2a + b)\,\alpha + (-3a + c) = 0.$$

We now have a linear equation in the variable $$\alpha$$. The number $$\alpha$$ is already known to be a root of the first quadratic. That quadratic,

$$x^2 + 2x + 3 = 0,$$

has discriminant

$$\Delta = 2^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 < 0,$$

so its two roots are non-real complex conjugates. Because the coefficients $$a, b, c$$ are required to be real, any complex root of the second quadratic must appear together with its conjugate. Therefore, if one root is common, its conjugate is automatically common as well, and the second quadratic shares both roots with the first. Consequently, the linear equation

$$(-2a + b)\,\alpha + (-3a + c) = 0$$

must be satisfied by both complex conjugate roots of the first quadratic. The only way a non-trivial linear expression can vanish for two distinct numbers is for each of its coefficients to be zero. Hence we must have

$$-2a + b = 0 \quad\text{and}\quad -3a + c = 0.$$

Solving these gives

$$b = 2a \qquad\text{and}\qquad c = 3a.$$

Writing the three coefficients in proportion, we obtain

$$a : b : c \;=\; a : 2a : 3a \;=\; 1 : 2 : 3.$$

Hence, the correct answer is Option C.