Adding the variances of 2 dependent variables and covariance

$$E(\hat \theta_1 ) = E(\hat \theta_2) = \theta_1$$ , $$Var (\hat \theta_1) = \sigma_1 , Var(\hat \theta_2) = \sigma_2, Cov(\hat \theta_1, \hat \theta_2) = \sigma_{12}$$

$$\hat \theta_3 = a \hat \theta_1 + (1-a) \hat \theta_2$$

The question asks me to find the value of a that minimizes the variance. I have the solutions to the answer but I don't understand how the covariance is used in calculating $$V(\hat \theta_3)$$

$$V(\hat \theta_3)$$ = $$V(a \hat \theta_1 + (1-a) \hat \theta_2 ) =$$ $$a^2 V( \hat \theta_1) + (1-a)^2 V(\hat \theta_2) + 2(1-a)Cov(\hat \theta_1, \hat \theta_2)$$

My issue isn't with determining the minimum value of alpha, that's just found by setting the derivative of $$V(\hat \theta_3)$$ to $$0$$ and checking 2nd deriviative for if it is a minimum, my issue is with the $$(1-a)$$

I do know that since both $$\hat \theta_1, \hat \theta_2$$ are not independent, there exists a covariance $$> 0$$, hence obviously in the calculations of adding 2 variances, you must take into account covariance and hence the theorem;

$$Var( X + Y ) =$$Var(X) + Var (Y) + 2 Cov(X,Y)

but where is the (1-a) coming from in $$2(1-a)Cov(\hat \theta_1, \hat \theta_2)$$ ?

\begin{align} V(a\hat{\theta}_1+(1-\theta_1)\hat{\theta_2})&=a^2V(\hat{\theta_1})+(1-a)^2V(\hat{\theta}_2)+2Cov(a\hat{\theta}_1, (1-a)\hat{\theta}_2)\\ &=a^2V(\hat{\theta_1})+(1-a)^2V(\hat{\theta}_2)+2a(1-a)Cov(\hat{\theta}_1, \hat{\theta}_2)\\ \end{align}
I believe it's a mistake, it should be $$a(1-a)$$.