smc

Since 3 has no factors of 2, $3^a$ will be odd for all values of $a$. Since $b$ is odd as well, $b-1$ must be even, so $(b-1{)}^2$ must be even. This means that for all choices of $c$, $(b-1{)}^{2}c$ must be even because any integer times an even number is still even. Since an odd number plus an even number is odd, $3^a+(b-1{)}^{2}c$ must be odd for all choices of $c$, which corresponds to answer choice $\boxed{\text{odd for all choices of c}}$.