Answer:

And the expected value for
a vector of zeros and the covariance matrix is given by:

So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:
The regression model assumes the errors are normally distributed.
Step-by-step explanation:
Assuming that we have n observations from a dependent variable Y , given by 
And for each observation of Y we have an independent variable X, given by 
We can write a linear model on this way:

Where
i a matrix for the error random variables, and for this case we can find the error ter like this:

And the expected value for
a vector of zeros and the covariance matrix is given by:

So we can see that the error terms not have a variance of 0. We can't assume that the errors are assumed to have an increasing mean, and we other property is that the errors are assumed independent and following a normal distribution so then the best option for this case would be:
The regression model assumes the errors are normally distributed.