For $n=1$ one can easily check that $2X_a+X_b$ has the binomial distribution on $\{0,1,2\}$ with success probability $\theta$. Then the expectation and the variance easily follow for any $n$. More precisely, one also deduces that $2X_a+X_b$ has the binomial distribution on $\{0,\ldots,2n\}$ with success probability $\theta$.

For $n=1$ one can easily check that $2X_a+X_b$ has the binomial distribution on $\{0,1,2\}$ with success probability $\theta$. Then the expectation and the variance easily follow for any $n$. More precisely, one also deduces that $2X_a+X_b$ has the binomial distribution on $\{0,\ldots,2n\}$ with success probability $\theta$.)

In fact, you can see the model as the one for the following experiment. Let $Y_i \sim Bin(2, \theta)$ modeling one observation. Then you observe $f(Y_i)$ with $f(0)=c, f(1)=b, f(2)=a$. The sum $Y_1 + \cdots + Y_n \sim Bin(2n,\theta)$ is a sufficient statistic and the mle of $\theta$ is $\frac{Y_1 + \cdots + Y_n}{2n}$, and $Y_1 + \cdots + Y_n = 0 \times \#c + 1\times \#b + 2\times \#a$.