How will I find them individually?

Clearly $a_0 = 1,$ and $a_1 = 3 \cdot 20 = 60.$ There are two ways to get $x^2$ in the expansion: two copies of $3x$ and eighteen of $1,$ or one copy of $4x^2$ and nineteen of $1.$ The former adds up to $9\binom{20}{2}x^2$ and the latter adds up to $4 \cdot 20x^2,$ so the sum is $1790x^2.$ Thus the answer is $1+60+1790 = \fbox{1851}.$

(If you wanted, you could also use Taylor series.)

$\begin{aligned} \left(1+3x+4x^2\right)^{20} & = \left(1+x(3+4x) \right)^{20} \\ & = 1 + 20x(3+4x)+ \frac {20\times 19}2 x^2(3+4x)^2 + \cdots + x^{20}(3+4x)^{20} \\ & = 1 + 60x+80x^2+ 190 x^2(9+24x+16x^2) + \cdots + x^{20}(3+4x)^{20} \\ & = 1 + 60x+80x^2+ 1710 x^2 + \cdots + x^{20}(3+4x)^{20} \\ & = 1 + 60x + 1790 x^2 + \cdots + x^{20}(3+4x)^{20} \end{aligned}$

$\implies a_0+a_1+a_2 = 1 + 60 + 1790 = \boxed{1851}$

Best solution for this will be like this.

Setting x=0, we get $1=a_{0}$ .

So $a_{0}=1$ .

Now for $a_{1}$ , differentiating the expression,

$\dfrac{d(1+3x+4x^2)^{20}}{dx}=0+a_{1}+2a_{2}x+...+na_{n}x^{n-1}$

$\implies 20.(1+3x+4x^2)^{19}.(3+8x)=a_{1}+2a_{2}x+..na_{n}x^{n-1}$ .

Set $x=0$ for the expression we got,

$\implies 20.(1)^{19}.3=a_{1}$

$\implies a_{1}=60$ .

Again differentiating the differentiated expression, we will get,

$20[19.(1+3x+4x^{2})^{18}.(3+8x)+8.(1+3x+4x^{2})^{19}=0+0+2a_{2}+3.2a_{3}x+...$

Setting x=0, and then solving for $a_{2}$ we get $a_{2}=1790$ .

So $\boxed{\boxed{\boxed{\boxed{a_{0}+a_{1}+a_{2}=1851}}}}$