$2b=a+c\Rightarrow$sum of $a \text{\&} c$ is even and $b$ depends on $a \text{\&} c$

Thus, we can not choose all three independently, but we can choose $a \text{\&} c$

Moreover, $a \text{and} c$ should be both either even or odd as their sum is even.

Out of $11$ consecutive natural numbers, either five are even and six are odd or six are even and five are odd.

Hence, $P=\frac{1}{2}\frac{{}^{5}C_2+{}^{6}C_2}{{}^{11}C_3}+\frac{1}{2}\frac{{}^{6}C_2+{}^{5}C_2}{{}^{11}C_3}=\frac{5}{33}$