Dummy coding of linear regression, intercept and constraint

Let the following multilevel problem, where we try to predict the credit card balance of individuals $$y_i$$:

$$x_{i 1}= \begin{cases}1 & \text { if } i \text { th person is from the South } \\ 0 & \text { if } i \text { th person is not from the South },\end{cases}$$

and the second should be

$$x_{i 2}= \begin{cases}1 & \text { if } i \text { th person is from the West } \\ 0 & \text { if } i \text { th person is not from the West. }\end{cases}$$

Then both of these variables can be used in the regression equation, in order to obtain the model $$y_i=\beta_0+\beta_1 x_{i 1}+\beta_2 x_{i 2}+\epsilon_i= \begin{cases}\beta_0+\beta_1+\epsilon_i & \text { if } i \text { th person is from the South } \\ \beta_0+\beta_2+\epsilon_i & \text { if } i \text { th person is from the West } \\ \beta_0+\epsilon_i & \text { if } i \text { th person is from the East. }\end{cases}$$

My problem now is to solve by linear regression under the problem that adding the intercept will add a lot of colinearity between the column vectors. One solution is to add one constraint:

$$\beta_0 + \beta_1 + \beta_2 = 0$$

And use restricted least square. But what is the interpretation of adding this constraint ? (Outside of the technical reason) I suppose that we cannot say anymore that $$\beta_0$$ can be interpreted as the average credit card balance for individuals from the east

$$y_i=\beta_0+\beta_1 x_{i 1}+\beta_2 x_{i 2}+\epsilon_i= \begin{cases}-\beta_2+\epsilon_i & \text { if } i \text { th person is from the South } \\ -\beta_1+\epsilon_i & \text { if } i \text { th person is from the West } \\ -\beta_1 - \beta_2+\epsilon_i & \text { if } i \text { th person is from the East. }\end{cases}$$

If you use a single column for each dummy variable, this wouldn't induce collinearity. Collinearity arises when for $$n$$ categories you use precisely $$n$$ columns (one-hot way) so that sum of these columns produces a column of ones, which is a multiple of intercept column. In your case there would be combinations like
0 1

Also note that imposing constraint $$\beta_0 + \beta_1 + \beta_2 = 0$$ and substituting for $$\beta_0$$ in the regression formula produces $$y_i = \beta_1(x_{i1} - 1) + \beta_2(x_{i2} - 1) + \epsilon_i$$ which amounts to re-coding categories as 0 and -1 and dropping intercept term.