Maximum principle for the heat equation involving an aditional linear term 

(a) Define $w=u \cdot e^{\alpha t}$. Then we have $w_t=e^{\alpha t} u_t + e^{\alpha t} \alpha u$ and $\Delta w=e^{\alpha t} \Delta u$. This implies that $$w_t-\Delta w = e^{\alpha t} \left( u_t + \alpha u -\Delta u \right) = e^{\alpha t} f \geq 0 \hspace{0.5cm} \text{in } \Omega_T.$$ Since $u=0 \Rightarrow w=0$ on $\Gamma _T$. Hence $w$ is a supersolution to the heat equation, and by Maximum principle (see Problem 14 in Chapter 2 of the PDE book by Evanse, first edition) the minimum of $w$ will occur on the boundary $\Gamma_t$. Therefore, $w \geq 0$ in $\Omega_T$. 

(b) In order to show $u_t \geq 0$, take the derivative with respect to $t$ and change the order of differentiation to get $$ \left( u_t \right) _t - \Delta \left( u_t \right) + \alpha \left(u_t\right) =0 \hspace{0.5cm} \text{in } \Omega_T.$$ In view of the argument in part (a), it is enpugh to show that $u_t \geq 0$ on $\Gamma_T$. Since $u=0$ on $\Gamma_T$, $u_t=0$ on $\Gamma_T \setminus \left(\Omega \times \{t=0\}\right)$. Furthermore, since $u \geq 0$ on $\Omega_T$,
$$u_t(x,0)=\lim_{t \rightarrow0}\frac{u(x,t)-u(x,0)}{t}=\lim_{t \rightarrow0}\frac{u(x,t)}{t}\geq 0.$$
Therefore $u_t \geq 0$ on $ \left(\Omega \times \{t=0\}\right)$, and hence $u_t\geq 0$ in $\Gamma_t$. By part (a) we conclude that $u_t \geq 0$ in $\overline{\Omega}_T $. 

P.S. I beilive in the staetement of the problem one needs to assume $u \in C^{2}_1(\overline{\Omega}_T)$ so we don't have to worry about existence of the limit above. I hope this helps.