# Is the coefficient of an orthogonal independent variable affected by the near-multicollinearity of two other independent variables?

Suppose there is a regression with three variables $$X_1, X_2, X_3$$. Say $$X_1$$ and $$X_2$$ have near-multicollinearity, but $$X_3$$ is (nearly) orthogonal to both.

Will $$\beta_3$$ experience the same problems the other two coefficients?

No, $$\beta_3$$ is not affected by the collinearity of $$X_1$$ and $$X_2$$. In fact, $$X_3$$ doesn't even have to be nearly orthogonal, as long as it is not collinear with $$X_1$$ and $$X_2$$.
First recall that notions like "collinearity" and "orthogonality" refer to the column vectors $$\mathbf x_1, \mathbf x_2, \mathbf x_3$$ of the design matrix $$D = (\mathbf x_1, \mathbf x_2, \mathbf x_3)$$. Now, let's call the response variable $$Y$$ and consider the situation where we have only two measurements. This is only so that it better corresponds to the figure, the arguments hold for an arbitrary number $$n$$ of measurements; see also the ending remark of this post.
In the figure, note that $$\mathbf x_1$$ and $$\mathbf x_2$$ are collinear, while $$\mathbf x_3$$ is neither collinear with $$\mathbf x_1$$ and $$\mathbf x_2$$ nor orthogonal to them. To approximate $$\mathbf y$$ with $$\mathbf x_1, \mathbf x_2$$, and $$\mathbf x_3$$, we have to create $$\mathbf z = \beta_1\mathbf x_1 + \beta_2 \mathbf x_2$$ with a linear combination of $$\mathbf x_1$$ and $$\mathbf x_2$$, and this $$\mathbf z$$ has to be such that, when adding an appropriate multiple of $$\mathbf x_3$$, we will reach $$\mathbf y$$: $$\beta \mathbf x_3 = \mathbf y - \mathbf z$$. This condition makes $$\mathbf z$$ unique.
Now, while there are, because of colinearity, infinitely many possibilities for $$\beta_1$$ and $$\beta_2$$ to create $$\mathbf z$$ from $$\mathbf x_1$$ and $$\mathbf x_2$$, there is only one multiple of $$\mathbf x_3$$ that, when added to $$\mathbf z$$, gives $$\mathbf y$$. Thus, $$\beta_3$$ is not affected by the collinearity.
If we have more than two measurements, it is probably the case that $$\mathbf y$$ will not be an element of the two-dimensional subplane $$\mathcal P$$ spanned by $$\mathbf x_1, \mathbf x_2$$, and $$\mathbf x_3$$. In this case, the above figure depicts this subplane $$\mathcal P$$ and the $$\mathbf y$$ in the figure will be the projection of the proper $$\mathbf y$$ to $$\mathcal P$$.