Given that $L\left(p,q\right)=\frac{ap+bq+c}{\sqrt{a^2+b^2}}$.

$L\left(\frac{2}{3},\frac{1}{3}\right)=\frac{a\left({\displaystyle \frac{2}{3}}\right)+b\left({\displaystyle \frac{1}{3}}\right)+c}{\sqrt{a^2+b^2}}=\frac{2a+b+3c}{3\sqrt{a^2+b^2}}$

$L\left(\frac{1}{3},\frac{2}{3}\right)=\frac{a\left({\displaystyle \frac{1}{3}}\right)+b\left({\displaystyle \frac{2}{3}}\right)+c}{\sqrt{a^2+b^2}}=\frac{a+2b+3c}{3\sqrt{a^2+b^2}}$

$L\left(2,2\right)=\frac{a\left(2\right)+b\left(2\right)+c}{\sqrt{a^2+b^2}}=\frac{2a+2b+c}{\sqrt{a^2+b^2}}=\frac{6a+6b+3c}{3\sqrt{a^2+b^2}}$

$L\left(\frac{2}{3},\frac{1}{3}\right)+L\left(\frac{1}{3},\frac{2}{3}\right)+L\left(2,2\right)$

$=\frac{2a+b+3c}{3\sqrt{a^2+b^2}}+\frac{a+2b+3c}{3\sqrt{a^2+b^2}}+\frac{6a+6b+3c}{3\sqrt{a^2+b^2}}$

$=\frac{9a+9b+9c}{3\sqrt{a^2+b^2}}=0$

$\therefore 9a+9b+9c=0\Rightarrow a+b+c=0 ...\left(1\right)$

By verifying the options

Substitute $(1,1)$ in the given equation of line,

$a\left(1\right)+b\left(1\right)+c=0 ...\left(2\right)$