Let ${\left(\mathrm{a}+\mathrm{bx}+{\mathrm{cx}}^2\right)}^{10}=\sum _{\mathrm{i}=0}^{20} {\mathrm{p}}_{\mathrm{i}}{\mathrm{x}}^{\mathrm{i}},\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{N}$. If p₁ = 20, and p₂ = 210 then 2(a + b + c) is equal

$$\begin{gathered} \text{ General term}\text{ is=}\frac{10!}{r_1!r_2!r_3!}\left(a{)}^{r_1}\right(bx{)}^{r_2}{\left(c{x}^2\right)}^{r_3} \\ \text{ For coeff. of }x:r_2+2r_3=1 \\ \begin{array}{l}{r}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{r}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{r}_3\\ 9\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array} \\ \therefore p_1=\text{ coef. of }x=\frac{10!}{9!}{a}^9{b}^1=20 \\ \Rightarrow a^9\cdot b=2 \end{gathered}$$
$\text{ Coff of }{x}^2:\frac{10!}{8!2!0!}{a}^8\cdot b^2+\frac{10!}{9!0!1!}\cdot a^9\cdot c=210$
$$\begin{gathered} \begin{array}{l}\Rightarrow 45a^8\cdot b^2+10\cdot a^9\cdot c=210\\ \Rightarrow 9a^8{b}^2+2a^9\cdot c=42\end{array} \\ \begin{array}{r}\therefore a=1,b=2,c=3\\ 2(a+b+c)=2(3+2+1)=12\end{array} \end{gathered}$$