Let $$(a + bx + cx^2)^{10} = \sum_{i=0}^{20} p_i x^i$$, $$a, b, c \in \mathbb{N}$$. If $$p_1 = 20$$ and $$p_2 = 210$$, then $$2(a + b + c)$$ is equal to

$$(a + bx + cx^2)^{10} = \sum p_i x^i$$

$$p_0 = a^{10}$$

For $$p_1$$: coefficient of $$x$$ comes from choosing $$bx$$ once and $$a$$ nine times:

$$p_1 = \binom{10}{1} a^9 \cdot b = 10a^9 b = 20$$

$$a^9 b = 2$$ ... (1)

Since $$a, b \in \mathbb{N}$$, the only solution to (1) is $$a = 1, b = 2$$.

For $$p_2$$: coefficient of $$x^2$$ comes from choosing $$bx$$ twice OR choosing $$cx^2$$ once:

$$p_2 = \binom{10}{2}a^8 b^2 + \binom{10}{1}a^9 c = 45 \times 1 \times 4 + 10 \times 1 \times c = 180 + 10c = 210$$

$$10c = 30$$, $$c = 3$$

$$2(a + b + c) = 2(1 + 2 + 3) = 12$$

This matches option 3: 12.