Let $f(x)=\left(1+x+x^2\right)^{10}=\sum_{r=0}^{20} a_r x^r$

The sum of odd coefficients: $S_{\text {odd }}=a_1+a_3+a_5+\cdots$ $+a_{19}$

Subtracting $11 a_2$ from above will give the answer

$$\begin{aligned} & S_{\text {odd }}=\frac{f(1)-f(-1)}{2} \\ & f(1)=(1+1+1)^{10}=3^{10} \\ & f(-1)=(1-1+1)^{10}=(1)^{10}=1 \\ & S_{\text {odd }}=\sum_{\text {odd } r} a_r=\frac{3^{10}-1}{2} \end{aligned}$$

Now for $a_2$

$$1+x+x^2=\frac{1-x^3}{1-x} \Rightarrow f(x)=\left(\frac{1-x^3}{1-x}\right)=\frac{\left(1-x^3\right)^{10}}{(1-x)^{10}}$$

Now use:

$$\begin{aligned} & \left(1-x^3\right)^{10}=\sum_{x=0}^{10}(-1)^k\binom{10}{k} x^{3 k} \\ & (1-x)^{-10}=\sum_{r=0}^{\infty}\binom{r+9}{9} x^r \end{aligned}$$

So

$$f(x)=\left(\sum_{k=0}^{10}(-1)^k\binom{10}{k} x^{3 k}\right) \cdot\left(\sum_{r=0}^{\infty}\binom{r+9}{9} x^r\right)$$

Only the term with $x^0$ from the first sum (i.e., $k=0$ ) can contribute to $x^2$, since all other $k \geq 1$ gives $x^{3 k} \geq$ $x^3$

From $\left(1-x^3\right)^{10}$ : the $x^0$ term is $\binom{10}{0}=1$

From $(1-x)^{-10}$ : the coefficient of $x^2$ is

$$\binom{2+9}{9}=\binom{11}{9}=55$$

Hence, $a_2=1.55=55$

$$\begin{aligned} & \text { Now, } S_{\text {odd }}-11 a_2=\frac{3^{10}-1}{2}-11 \cdot 55=121 k \\ & 3^{10}=59049 \end{aligned}$$

So:

$$\begin{aligned} & S=\frac{59049-1}{2}-605=\frac{59048}{2}-605 \\ & =29524-605=28919 \end{aligned}$$

So:

$$121 k=28919 \Rightarrow k=\frac{28919}{121}=239$$